JEE Circles Questions

The circle passes through $$(4,2)$$ and $$(0,2)$$, so the midpoint of these points lies on the perpendicular bisector.

Midpoint = $$(2, 2)$$. The line joining these points is horizontal ($$y = 2$$), so the perpendicular bisector is the vertical line $$x = 2$$.

The centre lies on both $$x = 2$$ and $$3x + 2y + 2 = 0$$:

$$3(2) + 2y + 2 = 0 \Rightarrow 2y = -8 \Rightarrow y = -4$$

Centre $$= (2, -4)$$.

Radius = distance from centre to $$(4,2)$$:

$$r = \sqrt{(4-2)^2 + (2+4)^2} = \sqrt{4+36} = \sqrt{40} = 2\sqrt{10}$$

For the chord with midpoint $$(1, 2)$$:

Distance from centre $$(2, -4)$$ to midpoint $$(1, 2)$$:

$$d = \sqrt{(2-1)^2 + (-4-2)^2} = \sqrt{1+36} = \sqrt{37}$$

Half-length of chord = $$\sqrt{r^2 - d^2} = \sqrt{40 - 37} = \sqrt{3}$$

Length of chord = $$2\sqrt{3}$$.

The correct answer is Option 3: $$2\sqrt{3}$$.

We have the circle $$x^2 + y^2 = 4$$ (centre O at origin, radius $$r = 2$$) and the chord $$x + y = 1$$.

The perpendicular from the centre (0, 0) to the line $$x + y = 1$$ meets the chord at its midpoint M.

Distance from origin to the line $$x + y - 1 = 0$$:

The midpoint M lies on $$x + y = 1$$ and on the line $$y = x$$ (perpendicular from origin to $$x + y = 1$$ has slope 1, since the chord has slope -1). So $$2x = 1$$, giving $$M = \left(\frac{1}{2}, \frac{1}{2}\right)$$.

Half-length of AB: $$\sqrt{r^2 - d^2} = \sqrt{4 - \frac{1}{2}} = \sqrt{\frac{7}{2}}$$.

So $$AB = 2\sqrt{\frac{7}{2}} = \sqrt{14}$$.

CD is the chord along the line $$y = x$$ (the perpendicular bisector of AB passing through M). This line passes through the centre O(0,0), so CD is a diameter.

Therefore $$CD = 2r = 4$$.

The quadrilateral ADBC has diagonals AB and CD. These two chords are perpendicular to each other (AB has slope -1, CD has slope 1), and they intersect at M.

For a quadrilateral whose diagonals are perpendicular and intersect at a point, the area is:

where $$d_1$$ and $$d_2$$ are the full diagonal lengths.

The answer is $$\boxed{2\sqrt{14}}$$, which corresponds to Option 3.

We are asked to find the area of the larger portion bounded between the circle $$x^2 + y^2 = 25$$ and the curve $$y = |x - 1|$$, express it as $$\frac{1}{4}(b\pi + c)$$ with $$b,c\in\mathbb{N}$$, and then determine $$b+c$$.

The circle centered at $$(0,0)$$ has radius 5, and the curve $$y = |x - 1|$$ is a V-shaped graph with vertex at $$(1,0)$$, lying entirely in the upper half-plane. To find their intersection points, substitute $$y = |x-1|$$ into $$x^2 + y^2 = 25$$, which gives

$$x^2 + (x-1)^2 = 25 \implies 2x^2 - 2x + 1 = 25 \implies 2x^2 - 2x - 24 = 0 \implies x^2 - x - 12 = 0 \implies (x - 4)(x + 3) = 0,$$

so $$x = 4$$ or $$x = -3$$. The intersection points are therefore $$(4,3)$$ and $$(-3,4)$$.

The smaller region is bounded above by the upper semicircle and below by the V-curve from $$x = -3$$ to $$x = 4$$. Its area is

$$A_{\text{small}} = \int_{-3}^{4}\sqrt{25 - x^2}\,dx \;-\;\int_{-3}^{4}|x-1|\,dx.$$

First, using the standard antiderivative $$\int \sqrt{a^2 - x^2}\,dx = \frac{x\sqrt{a^2 - x^2}}{2} + \frac{a^2}{2}\sin^{-1}\frac{x}{a} + C$$ with $$a=5$$, we evaluate at the bounds:

At $$x=4$$: $$\frac{4\cdot3}{2} + \frac{25}{2}\sin^{-1}\!\frac{4}{5} = 6 + \frac{25}{2}\sin^{-1}\!\frac{4}{5},$$

At $$x=-3$$: $$\frac{-3\cdot4}{2} + \frac{25}{2}\sin^{-1}\!\frac{-3}{5} = -6 - \frac{25}{2}\sin^{-1}\!\frac{3}{5}.$$

Since $$\sin^{-1}\!\frac{4}{5} + \sin^{-1}\!\frac{3}{5} = \frac{\pi}{2}$$, it follows that

$$\int_{-3}^{4}\sqrt{25 - x^2}\,dx = \bigl(6 + \tfrac{25}{2}\sin^{-1}(4/5)\bigr) - \bigl(-6 - \tfrac{25}{2}\sin^{-1}(3/5)\bigr) = 12 + \frac{25\pi}{4}.$$

Next, split the integral of the V-curve at $$x=1$$:

$$\int_{-3}^{1}(1-x)\,dx = \Bigl[x - \tfrac{x^2}{2}\Bigr]_{-3}^{1} = \frac12 + \frac{15}{2} = 8,$$

$$\int_{1}^{4}(x-1)\,dx = \Bigl[\tfrac{x^2}{2} - x\Bigr]_{1}^{4} = (8 - 4) - \bigl(\tfrac12 - 1\bigr) = 4 + \tfrac12 = \frac{9}{2},$$

so that $$\int_{-3}^{4}|x-1|\,dx = 8 + \frac{9}{2} = \frac{25}{2}.$$

Therefore,

$$A_{\text{small}} = \Bigl(12 + \frac{25\pi}{4}\Bigr) - \frac{25}{2} = \frac{25\pi - 2}{4}.$$

Since the total area of the circle is $$25\pi$$, the larger region has area

$$A_{\text{larger}} = 25\pi - \frac{25\pi - 2}{4} = \frac{100\pi - 25\pi + 2}{4} = \frac{75\pi + 2}{4} = \frac{1}{4}(75\pi + 2).$$

Thus, $$b = 75$$ and $$c = 2$$, giving $$b + c = 77$$.

The circle is tangent to the x-axis at the point $$\left(a,0\right)$$ with $$a \lt 0$$.
If a circle touches the x-axis, its centre must be exactly $$r$$ units above the point of contact.
Hence the centre of the required circle is $$C\left(a,r\right)$$ and its radius is $$r$$.

The circle is also tangent to the parabola $$y^{2}=9x$$ at the point $$P(4,6)$$.
Therefore two conditions hold:

1. CP = r   (distance condition)

2. CP is perpendicular to the tangent to the parabola at P   (normal condition)

Step 1: Distance condition

The distance CP is

$$CP=\sqrt{(4-a)^{2}+(6-r)^{2}}$$

Since CP = r, we get

$$(4-a)^{2}+(6-r)^{2}=r^{2} \qquad -(1)$$

Step 2: Normal condition

For the parabola $$y^{2}=9x$$, differentiate:
$$2y\,\frac{dy}{dx}=9 \;\;\Rightarrow\;\; \frac{dy}{dx}=\frac{9}{2y}$$

At P(4,6):
$$m_{\text{tangent}} = \frac{9}{2\cdot6} = \frac{3}{4}$$

Hence the slope of the normal is the negative reciprocal:

$$m_{\text{normal}} = -\frac{4}{3}$$

Equation of the normal through P:

$$y-6 = -\frac{4}{3}\,(x-4)$$

Since C(a,r) lies on this normal, substitute:

$$r-6 = -\frac{4}{3}\,(a-4)$$

Solve for r:

$$r = 6 - \frac{4}{3}(a-4)=6-\frac{4a}{3}+\frac{16}{3} = \frac{18}{3}+\frac{16}{3}-\frac{4a}{3} = \frac{34-4a}{3} \qquad -(2)$$

Step 3: Substitute into the distance equation

First compute $$6-r$$ using (2):
$$6-r = 6-\left(\frac{34-4a}{3}\right) = \frac{18}{3}-\frac{34-4a}{3} = \frac{18-34+4a}{3} = \frac{4a-16}{3} = \frac{4(a-4)}{3}$$

Put this in equation (1):

$$(4-a)^{2} + \left(\frac{4(a-4)}{3}\right)^{2} = \left(\frac{34-4a}{3}\right)^{2}$$

Multiply by 9 to clear denominators:

$$9(4-a)^{2} + 16(a-4)^{2} = (4a-34)^{2}$$

Because $$(a-4)^{2}=(4-a)^{2}$$, set $$s=(4-a)^{2}$$. Then

$$9s + 16s = 25s = (4a-34)^{2}$$

Step 4: Solve for a

Take square roots (all quantities are non-negative):

$$5|4-a| = |4a-34|$$

Since $$a \lt 0$$, $$4-a \gt 0$$ and $$4a-34 \lt 0$$, so

$$5(4-a) = -(4a-34) = 34-4a$$

$$20 - 5a = 34 - 4a$$

$$-a = 14 \quad\Longrightarrow\quad a = -14$$

Step 5: Find r

Insert $$a=-14$$ into (2):

$$r = \frac{34 - 4(-14)}{3} = \frac{34 + 56}{3} = \frac{90}{3} = 30$$

Hence the required radius is $$\mathbf{30}$$.

Let the two given lines be $$L_1 : x + y = 3$$ and $$L_2 : x - y = 3$$.

Step 1 - Find their intersection and show they are perpendicular.
Solving the two equations simultaneously gives the point of intersection $$I(3,0)$$.
The slopes are $$-1$$ for $$L_1$$ and $$+1$$ for $$L_2$$ and their product is $$-1$$, hence the lines are perpendicular.

Step 2 - Locate the possible centres.
For a circle to be tangent to two perpendicular lines, its centre must lie on one of the two angle-bisectors of the lines.
The angle-bisectors of the pair $$Y = \pm X$$ (after shifting the origin to $$I(3,0)$$ by putting $$X = x-3,\; Y = y$$) are
  • $$Y = 0 \; \Longrightarrow\; y = 0$$, a horizontal line through $$I$$, and
  • $$X = 0 \; \Longrightarrow\; x = 3$$, a vertical line through $$I$$.

We therefore examine two cases.

Case 1: Centre $$C(h,0)$$ lies on $$y = 0$$.

Radius as distance from the centre to either line:
$$r = \dfrac{|h-3|}{\sqrt2} \; \Longrightarrow\; r^{2} = \dfrac{(h-3)^{2}}{2}$$.

Radius as distance from the centre to the fixed point $$P(-9,4)$$:
$$r^{2} = (h+9)^{2} + (0-4)^{2} = (h+9)^{2} + 16$$.

Equating the two expressions for $$r^{2}$$ gives
$$\dfrac{(h-3)^{2}}{2} = (h+9)^{2} + 16$$.
Multiplying by $$2$$ and simplifying:
$$(h-3)^{2} = 2(h+9)^{2} + 32$$
$$h^{2} - 6h + 9 = 2h^{2} + 36h + 194$$
$$0 = h^{2} + 42h + 185$$.

Solving the quadratic:
$$h = \dfrac{-42 \pm \sqrt{42^{2} - 4\cdot1\cdot185}}{2} = \dfrac{-42 \pm 32}{2}$$
$$\Longrightarrow\; h = -5 \quad\text{or}\quad h = -37$$.

Thus two centres are $$C_1(-5,0)$$ and $$C_2(-37,0)$$.
Corresponding radii:
$$r_1^{2} = ( -5 + 9)^{2} + 16 = 4^{2} + 16 = 32 \;\; \Longrightarrow\;\; r_1 = 4\sqrt2$$,
$$r_2^{2} = ( -37 + 9)^{2} + 16 = 28^{2} + 16 = 800 \;\; \Longrightarrow\;\; r_2 = 20\sqrt2$$.

Case 2: Centre $$C(3,k)$$ lies on $$x = 3$$.

Radius from the centre to either line:
$$r = \dfrac{|k|}{\sqrt2} \; \Longrightarrow\; r^{2} = \dfrac{k^{2}}{2}$$.

Radius from the centre to the fixed point $$P(-9,4)$$:
$$r^{2} = (3 + 9)^{2} + (k - 4)^{2} = 144 + (k - 4)^{2}$$.

Equating gives
$$\dfrac{k^{2}}{2} = 144 + (k - 4)^{2}$$
$$k^{2} = 288 + 2(k^{2} - 8k + 16)$$
$$k^{2} = 2k^{2} - 16k + 320$$
$$0 = k^{2} - 16k + 320$$.

The discriminant of this quadratic is
$$\Delta = (-16)^{2} - 4\cdot1\cdot320 = 256 - 1280 = -1024 \lt 0,$$
so there is no real value of $$k$$. Consequently, no circle has its centre on $$x = 3$$.

Step 3 - Compute the required absolute difference.
The only admissible radii are $$r_1 = 4\sqrt2$$ and $$r_2 = 20\sqrt2$$.
Therefore
$$|\,r_2^{2} - r_1^{2}\,| = |\,800 - 32\,| = 768.$$

Hence, the absolute difference between the squares of the radii is 768.

The general equation of a circle in the plane is $$x^{2}+y^{2}+2gx+2fy+c=0$$, where

center = $$(-g,\,-f)$$ and radius $$r=\sqrt{\,g^{2}+f^{2}-c\,}$$.

Because three non-collinear points determine a unique circle, first obtain the circle passing through the points $$(4,6)$$, $$(-1,5)$$ and $$(0,0)$$.

Substituting $$(4,6)$$ gives
$$16+36+8g+12f+c=0 \quad\Rightarrow\quad 52+8g+12f+c=0 \; -(1)$$

Substituting $$(-1,5)$$ gives
$$1+25-2g+10f+c=0 \quad\Rightarrow\quad 26-2g+10f+c=0 \; -(2)$$

Substituting $$(0,0)$$ gives
$$c=0 \; -(3)$$

With $$c=0$$, equations $$(1)$$ and $$(2)$$ reduce to

$$52+8g+12f=0 \; -(4)$$
$$26-2g+10f=0 \; -(5)$$

Solve $$(4)$$ and $$(5)$$ simultaneously:

Multiply $$(5)$$ by $$4$$: $$104-8g+40f=0$$.
Add to $$(4)$$: $$ (52+104) + (12f+40f) = 0 \;\Rightarrow\; 156+52f=0$$,
so $$f=-3$$.

Insert $$f=-3$$ into $$(4)$$:
$$52+8g+12(-3)=0 \;\Rightarrow\; 52+8g-36=0 \;\Rightarrow\; 8g=-16 \;\Rightarrow\; g=-2$$.

Hence the circle through the three known points is
$$x^{2}+y^{2}-4x-6y=0$$.

Its center is $$(-g,\,-f)=(2,3)$$ and the radius is

$$r=\sqrt{g^{2}+f^{2}-c}=\sqrt{(-2)^{2}+(-3)^{2}-0}=\sqrt{4+9}= \sqrt{13}$$,
so $$r^{2}=13$$.

Now impose the condition that the fourth point $$(k,\,3k)$$ also satisfies the circle equation:

$$k^{2}+(3k)^{2}-4k-6(3k)=0$$
$$\Longrightarrow\; k^{2}+9k^{2}-4k-18k=0$$
$$\Longrightarrow\; 10k^{2}-22k=0$$
$$\Longrightarrow\; 2k\,(5k-11)=0$$.

Because the four points are distinct, $$k \neq 0$$; therefore

$$5k-11=0 \;\Rightarrow\; k=\frac{11}{5}$$.

Finally, evaluate $$10k+r^{2}$$:

$$10k+r^{2}=10\left(\frac{11}{5}\right)+13=22+13=35$$.

Hence $$10k+r^{2}=35$$, which corresponds to Option D.

We need to find the ordered pair $$(2a, b^2)$$ for a circle that touches the x-axis at $$(a, 0)$$ with $$a \gt 0$$, cuts an intercept of length $$b$$ on the y-axis, and lies below the x-axis.

A circle touching the x-axis at $$(a, 0)$$ has its centre at $$(a, r)$$ where $$r$$ is the radius. Since the circle lies below the x-axis, the centre is at $$(a, -r)$$ and the radius is $$r$$. Its equation is

$$(x - a)^2 + (y + r)^2 = r^2$$

Expanding this gives

$$x^2 - 2ax + a^2 + y^2 + 2ry + r^2 = r^2$$

which simplifies to

$$x^2 + y^2 - 2ax + 2ry + a^2 = 0$$

Comparing this with the form $$x^2 + y^2 - \alpha x + \beta y + \gamma = 0$$, we identify $$\alpha = 2a$$, $$\beta = 2r$$, and $$\gamma = a^2$$. Thus, the first element of the ordered pair is $$2a = \alpha$$.

To find the y-axis intercept, set $$x = 0$$ in the circle’s equation, yielding

$$y^2 + \beta y + \gamma = 0$$

The intercept of length $$b$$ on the y-axis satisfies

$$b = \sqrt{\beta^2 - 4\gamma}$$

so that

$$b^2 = \beta^2 - 4\gamma$$

Hence, the ordered pair $$(2a, b^2)$$ becomes

$$ (2a, b^2) = (\alpha, \beta^2 - 4\gamma)$$

The correct answer is Option D: $$(\alpha, \beta^2 - 4\gamma)$$.

The given circle has centre $$(0,\alpha)$$ and radius $$r$$. The line $$y = 2x$$ is tangent to the circle.

Write the line in the form $$Ax + By + C = 0$$: $$y = 2x \;\; \Longrightarrow \;\; 2x - y = 0$$ so $$A = 2,\; B = -1,\; C = 0$$.

Distance of centre $$(0,\alpha)$$ from the line equals the radius: $$\frac{|2\cdot 0 + (-1)\alpha + 0|}{\sqrt{2^{2}+(-1)^{2}}} = r \;\; \Longrightarrow \;\; \frac{\alpha}{\sqrt{5}} = r \quad -(1)$$

Also given: $$\alpha + r = 5 + \sqrt{5} \quad -(2)$$

Substitute $$r = \alpha/\sqrt{5}$$ from $$(1)$$ into $$(2)$$: $$\alpha + \frac{\alpha}{\sqrt{5}} = 5 + \sqrt{5} \;\; \Longrightarrow \;\; \alpha\!\left(1 + \frac{1}{\sqrt{5}}\right) = 5 + \sqrt{5}$$

Multiply numerator and denominator by $$\sqrt{5}$$: $$\alpha = \frac{5 + \sqrt{5}}{1 + 1/\sqrt{5}} = \frac{5 + \sqrt{5}}{(\sqrt{5}+1)/\sqrt{5}} = \sqrt{5}\,\frac{5 + \sqrt{5}}{\sqrt{5}+1} = \frac{5\sqrt{5}+5}{\sqrt{5}+1} = 5$$

Therefore $$\alpha = 5$$ and from $$(1)$$ $$r = \frac{\alpha}{\sqrt{5}} = \frac{5}{\sqrt{5}} = \sqrt{5}$$

Next find the point of contact $$A_1(x_1,y_1)$$. For a tangent, the radius $$\overrightarrow{OA_1}$$ is normal to the line. A normal vector to $$2x - y = 0$$ is $$(2,-1)$$ whose length is $$\sqrt{5}$$. Unit normal $$\mathbf{u} = \left(\dfrac{2}{\sqrt{5}}, -\dfrac{1}{\sqrt{5}}\right)$$.

Hence $$A_1 = (0,\alpha) + r\,\mathbf{u} = \left(0 + \sqrt{5}\cdot\frac{2}{\sqrt{5}},\; 5 + \sqrt{5}\cdot\left(-\frac{1}{\sqrt{5}}\right)\right) = (2,4)$$

Since $$A_1B_1$$ is a diameter, the centre is the midpoint: $$\overrightarrow{OA_1} = (2,-1) \implies \overrightarrow{OB_1} = -(2,-1) = (-2,1)$$ so $$B_1 = (0,\alpha) + (-2,1) = (-2,6)$$

Collecting the results:
$$\alpha = 5,\quad r = \sqrt{5},\quad A_1 = (2,4),\quad B_1 = (-2,6)$$

Matching with List-II:
(P) $$\alpha$$ $$\to$$ 5 (entry 4)
(Q) $$r$$ $$\to$$ $$\sqrt{5}$$ (entry 2)
(R) $$A_1$$ $$\to$$ $$(2,4)$$ (entry 5)
(S) $$B_1$$ $$\to$$ $$(\!-2,6)$$ (entry 3)

Thus the correct option is:
Option C: (P) $$\to$$ (4), (Q) $$\to$$ (2), (R) $$\to$$ (5), (S) $$\to$$ (3)

We begin by finding the centre and radius of the circle whose equation is $$x^2 + y^2 - 10x + 4y + 13 = 0$$. Completing the square yields $$(x^2 - 10x + 25) + (y^2 + 4y + 4) + 13 - 25 - 4 = 0$$, which simplifies to $$(x - 5)^2 + (y + 2)^2 = 16$$. Hence, the centre is $$(5, -2)$$ and the radius is $$r = 4$$, making the length of its diameter $$8$$ and the half-length of the diameter (the chord length) equal to $$4$$.

The centre of the circle $$C$$ is determined by the intersection of the lines $$2x + 3y = 12$$ and $$3x - 2y = 5$$. Solving these simultaneously, multiplying the first by 2 gives $$4x + 6y = 24$$ and multiplying the second by 3 gives $$9x - 6y = 15$$. Adding these equations yields $$13x = 39$$, so $$x = 3$$. Substituting back into $$2x + 3y = 12$$ gives $$2(3) + 3y = 12$$, hence $$y = 2$$. Therefore, the centre of $$C$$ is $$(3, 2)$$.

The given diameter of the first circle serves as a chord of circle $$C$$ and its midpoint is the centre $$(5, -2)$$ of the first circle. The perpendicular distance $$d$$ from the centre of $$C$$ to this chord is found by the distance formula: $$d = \sqrt{(5 - 3)^2 + (-2 - 2)^2} = \sqrt{20} = 2\sqrt{5}\,. $$

In any circle, the relationship between the radius $$R$$, the half-length of a chord $$l$$, and the perpendicular distance $$d$$ from the centre to the chord is given by $$R^2 = d^2 + l^2$$. Substituting $$d^2 = 20$$ and $$l = 4$$ gives $$R^2 = 20 + 16 = 36$$, so $$R = 6$$.

Therefore, the radius of circle $$C$$ is 6.

The first circle is $$(x+1)^2 + (y+2)^2 = r^2$$.
Its centre is $$C_1(-1,-2)$$ and its radius is $$r$$.

Rewrite the second circle $$x^2 + y^2 - 4x - 4y + 4 = 0$$ in standard form.

Bring the constant to the right side:
$$x^2 - 4x + y^2 - 4y = -4$$

Complete the squares:
$$x^2 - 4x + 4 + y^2 - 4y + 4 = -4 + 4 + 4$$
$$\Rightarrow (x-2)^2 + (y-2)^2 = 4$$

Thus the second circle has centre $$C_2(2,2)$$ and radius $$R = 2$$.

Let $$d$$ be the distance between the centres.

$$d = \sqrt{(2+1)^2 + (2+2)^2} = \sqrt{3^2 + 4^2} = \sqrt{25} = 5$$

For two circles of radii $$r$$ and $$R$$ to intersect at exactly two distinct points, the distance between their centres must satisfy
$$|r - R| \lt d \lt r + R$$

Substitute $$R = 2$$ and $$d = 5$$:

Left inequality:
$$|r - 2| \lt 5 \quad\Longrightarrow\quad -5 \lt r - 2 \lt 5$$
$$\Rightarrow -3 \lt r \lt 7$$

Right inequality:
$$5 \lt r + 2 \quad\Longrightarrow\quad r \gt 3$$

Combine the two results:
$$3 \lt r \lt 7$$

Therefore $$r$$ must lie between $$3$$ and $$7$$ (exclusive). This matches Option C.

Answer - Option C: $$3 \lt r \lt 7$$

Circle $$C: x^2 + y^2 = 10$$. Chord $$PQ$$ on $$x + y = 2$$. Chord $$MN$$ length 2, slope $$-1$$.

Distance of $$PQ$$ from origin.

The line is $$x + y - 2 = 0$$. $$d_1 = \frac{|0+0-2|}{\sqrt{1^2+1^2}} = \frac{2}{\sqrt{2}} = \sqrt{2}$$.

 Distance of $$MN$$ from origin.

$$MN$$ has slope $$-1$$, so it is parallel to $$PQ$$ (equation $$x + y = k$$).

Length of chord $$L = 2\sqrt{r^2 - d_2^2}$$.

$$2 = 2\sqrt{10 - d_2^2} \implies 1 = 10 - d_2^2 \implies d_2 = \sqrt{9} = 3$$.

Distance between chords.

Since they are parallel and on the same side (or opposite), the distance is $$|d_2 - d_1|$$.

Distance = $$\mathbf{3 - \sqrt{2}}$$ (Option A).

We need to find $$m + n^2$$ where $$m$$ and $$n$$ are the area and perimeter of a square inscribed in a circle, which itself is inscribed in an equilateral triangle of side 12.

In an equilateral triangle of side $$a$$, the inradius is given by $$r = \frac{a}{2\sqrt{3}} = \frac{a\sqrt{3}}{6}$$, so for $$a = 12$$ we have $$r = \frac{12\sqrt{3}}{6} = 2\sqrt{3}$$.

A square inscribed in a circle has its diagonal equal to the diameter, so the diagonal of the square is $$2r = 4\sqrt{3}$$. Since for a square with side $$s$$ the diagonal is $$s\sqrt{2}$$, it follows that $$s\sqrt{2} = 4\sqrt{3} \implies s = \frac{4\sqrt{3}}{\sqrt{2}} = 2\sqrt{6}$$.

The area of the square is $$m = s^2 = (2\sqrt{6})^2 = 4 \times 6 = 24$$, and its perimeter is $$n = 4s = 4 \times 2\sqrt{6} = 8\sqrt{6}$$.

Since $$n^2 = (8\sqrt{6})^2 = 64 \times 6 = 384$$, we obtain $$m + n^2 = 24 + 384 = 408$$.

The correct answer is Option 1: 408.

The circle passes through $$(0, 0)$$, $$(1, 0)$$, $$(0, 1)$$, and $$(2k, 3k)$$.

The general equation of a circle through the origin: $$x^2 + y^2 + Dx + Ey = 0$$.

Using $$(1, 0)$$: $$1 + D = 0 \Rightarrow D = -1$$.

Using $$(0, 1)$$: $$1 + E = 0 \Rightarrow E = -1$$.

Circle: $$x^2 + y^2 - x - y = 0$$.

Using $$(2k, 3k)$$:

$$4k^2 + 9k^2 - 2k - 3k = 0$$

$$13k^2 - 5k = 0$$

$$k(13k - 5) = 0$$

Since the four points are distinct and $$k \neq 0$$ (otherwise $$(2k, 3k) = (0,0)$$):

$$k = \frac{5}{13}$$

The answer is $$\frac{5}{13}$$, which corresponds to Option (3).

Orthocentre P(6,1), A(5,-2), B(8,3). AP⊥BC: slope AP = (1+2)/(6-5) = 3. Slope BC = -1/3.

BP⊥AC: slope BP = (1-3)/(6-8) = 1. Slope AC = -1.

BC: passes through B(8,3) slope -1/3: y-3 = -1/3(x-8) → x+3y = 17.

AC: passes through A(5,-2) slope -1: y+2 = -(x-5) → x+y = 3.

C is intersection: x+3y=17, x+y=3 → 2y=14, y=7, x=-4. C(-4,7).

C lies on circle: x²+y² = 16+49 = 65.

The correct answer is Option (3): x²+y²-65=0.

A circle of radius 1 is touched by lines through $$(3,2)$$ parallel to the coordinate axes; the one closest to the origin is sought. The lines through $$(3,2)$$ parallel to the axes are $$x = 3$$ and $$y = 2$$, and since the circle of radius 1 must touch both lines the center must be at distance 1 from each. Thus the possible centers are $$(3 \pm 1, 2 \pm 1) = (2,1), (2,3), (4,1), (4,3)$$.

To be closest to the origin the chosen center must minimize its distance from $$(0,0)$$:
- $$(2,1)$$: distance = $$\sqrt{5} \approx 2.24$$
- $$(2,3)$$: distance = $$\sqrt{13} \approx 3.61$$
- $$(4,1)$$: distance = $$\sqrt{17} \approx 4.12$$
- $$(4,3)$$: distance = $$5$$. The minimum occurs at $$(2,1)$$ as the center.

The shortest distance from this circle to the point $$(5,5)$$ is given by $$d = \text{distance from center to point} - r = \sqrt{(5-2)^2 + (5-1)^2} - 1 = \sqrt{9+16} - 1 = 5 - 1 = 4$$

The correct answer is Option (3): 4.

Find intersection of $$3x + 5y = 1$$ and $$(2+c)x + 5c^2y = 1$$.

At $$c = 1$$: both become $$3x + 5y = 1$$, so we need limits.

Subtracting: $$(2+c)x + 5c^2y - 3x - 5y = 0 \implies (c-1)x + 5(c^2-1)y = 0$$.

$$(c-1)x + 5(c-1)(c+1)y = 0$$. For $$c \neq 1$$: $$x + 5(c+1)y = 0$$.

As $$c \to 1$$: $$x + 10y = 0 \implies x = -10y$$.

Substituting into $$3x + 5y = 1$$: $$-30y + 5y = 1 \implies -25y = 1 \implies y = -\frac{1}{25}$$.

$$x = -10 \times (-\frac{1}{25}) = \frac{2}{5}$$.

So $$h = \frac{2}{5}$$, $$k = -\frac{1}{25}$$.

Radius = distance from centre $$(h,k)$$ to $$(2,0)$$:

$$r^2 = (2-\frac{2}{5})^2 + (0+\frac{1}{25})^2 = (\frac{8}{5})^2 + (\frac{1}{25})^2 = \frac{64}{25} + \frac{1}{625} = \frac{1600+1}{625} = \frac{1601}{625}$$

Circle equation: $$(x-\frac{2}{5})^2 + (y+\frac{1}{25})^2 = \frac{1601}{625}$$

Expanding: $$x^2 - \frac{4x}{5} + \frac{4}{25} + y^2 + \frac{2y}{25} + \frac{1}{625} = \frac{1601}{625}$$

$$x^2 + y^2 - \frac{4x}{5} + \frac{2y}{25} = \frac{1601}{625} - \frac{4}{25} - \frac{1}{625} = \frac{1601 - 100 - 1}{625} = \frac{1500}{625} = \frac{12}{5}$$

Multiplying by 25: $$25x^2 + 25y^2 - 20x + 2y = 60$$

$$25x^2 + 25y^2 - 20x + 2y - 60 = 0$$.

The correct answer is Option 4.

Let $$ABCD$$ be a square of side 4 and $$AEFG$$ be a square of side 2, where $$E$$ is on $$AB$$ and $$F$$ is on diagonal $$AC$$.

Place $$A=(0,0)$$, $$B=(4,0)$$, $$C=(4,4)$$, and $$D=(0,4)$$. Since $$E$$ is on $$AB$$ with $$AE=2$$, it follows that $$E=(2,0)$$, and the diagonal $$AC$$ has equation $$y=x$$.

Square $$AEFG$$ of side 2 has vertices in order: starting from $$A=(0,0)$$ and $$E=(2,0)$$, going counterclockwise gives $$F=(2,2)$$ and $$G=(0,2)$$. One can verify that $$EF=FG=GA=AE=2$$ and that $$F(2,2)$$ lies on $$y=x$$.

We seek a circle through $$F$$ tangent to lines $$BC: x=4$$ and $$CD: y=4$$. Since these lines are perpendicular, the center must be equidistant from both, so it is $$(4-r,4-r)$$ with radius $$r$$.

Imposing that $$F(2,2)$$ lies on the circle gives

$$ (2 - (4 - r))^2 + (2 - (4 - r))^2 = r^2 $$

$$ 2(r - 2)^2 = r^2 $$

$$ 2r^2 - 8r + 8 = r^2 $$

$$ r^2 - 8r + 8 = 0 $$

The answer is Option D: $$r^2 - 8r + 8 = 0$$.

Two circles intersect at two distinct points if $$|r_1 - r_2| < d < r_1 + r_2$$, where $$d$$ is the distance between centers.

Circle Properties:

o $$C$$: Center $$C_1(0,0)$$, radius $$r_1 = 2$$.

o $$C'$$: Center $$C_2(2\lambda, 0)$$, radius $$r_2 = \sqrt{(2\lambda)^2 - 9} = \sqrt{4\lambda^2 - 9}$$.

For intersection:

$$d = \sqrt{(2\lambda - 0)^2 + 0} = |2\lambda|$$.

We need $$r_2$$ to be real: $$4\lambda^2 - 9 > 0 \implies |\lambda| > 3/2$$.

Condition: $$|r_1 - r_2| < d < r_1 + r_2 \implies (r_1 - r_2)^2 < d^2 < (r_1 + r_2)^2$$.

$$d^2 < (r_1 + r_2)^2 \implies 4\lambda^2 < (2 + \sqrt{4\lambda^2 - 9})^2$$

$$4\lambda^2 < 4 + (4\lambda^2 - 9) + 4\sqrt{4\lambda^2 - 9} \implies 5 < 4\sqrt{4\lambda^2 - 9}$$

$$25 < 16(4\lambda^2 - 9) \implies 25 < 64\lambda^2 - 144 \implies 64\lambda^2 > 169 \implies |\lambda| > 13/8$$.

Also check $$(r_1 - r_2)^2 < d^2$$, which yields the same result.

The range is $$|\lambda| > 13/8$$, which is $$\mathbb{R} - [-13/8, 13/8]$$.

So, $$a = -13/8$$ and $$b = 13/8$$.

$$x = 8a + 12 = 8(-13/8) + 12 = -13 + 12 = -1$$.

$$y = 16b - 20 = 16(13/8) - 20 = 26 - 20 = 6$$.

Point $$(x, y) = (-1, 6)$$

Plug $$(-1, 6)$$ into $$6x^2 + y^2 = 42$$: $$6(-1)^2 + (6)^2 = 6 + 36 = 42$$.

The point lies on $$6x^2 + y^2 = 42$$.

The centres of the two circles are
$$C_1(\alpha ,\beta) \quad\text{and}\quad C_2\left(8,\frac{15}{2}\right).$$

Given that the point $$P(6,6)$$ divides the line segment $$C_1C_2$$ internally in the ratio $$2:1$$, we use the section (internal‐division) formula.
If $$P(x_P,y_P)$$ divides $$A(x_A,y_A)$$ and $$B(x_B,y_B)$$ so that $$AP:PB = m:n$$, then
$$x_P = \frac{n\,x_A + m\,x_B}{m+n}, \qquad y_P = \frac{n\,y_A + m\,y_B}{m+n}.$$ Here $$m=2,\; n=1,\; A\equiv C_1(\alpha,\beta),\; B\equiv C_2\left(8,\frac{15}{2}\right).$$

Applying the formula to the $$x$$-coordinate:
$$6 = x_P = \frac{1\cdot \alpha + 2\cdot 8}{2+1} = \frac{\alpha + 16}{3}$$ $$\Rightarrow \alpha + 16 = 18 \;\Rightarrow\; \alpha = 2.$$

Applying the formula to the $$y$$-coordinate:
$$6 = y_P = \frac{1\cdot \beta + 2\left(\dfrac{15}{2}\right)}{3} = \frac{\beta + 15}{3}$$ $$\Rightarrow \beta + 15 = 18 \;\Rightarrow\; \beta = 3.$$

Thus
$$C_1(2,3),\quad C_2\left(8,\frac{15}{2}\right).$$

The point $$P(6,6)$$ is the common point of contact, so it lies on both circles:

Radius of $$C_1$$:
$$r_1 = \sqrt{(6-2)^2 + (6-3)^2} = \sqrt{4^2 + 3^2} = \sqrt{25} = 5,$$ $$\therefore\; r_1^{2}=25.$$

Radius of $$C_2$$:
$$r_2 = \sqrt{(6-8)^2 + \left(6-\frac{15}{2}\right)^2} = \sqrt{(-2)^2 + \left(-\frac{3}{2}\right)^2} = \sqrt{4 + \frac{9}{4}} = \sqrt{\frac{25}{4}} = \frac{5}{2},$$ $$\therefore\; r_2^{2} = \frac{25}{4}.$$

Finally, evaluate the required expression:
$$(\alpha + \beta) + 4\left(r_1^{2} + r_2^{2}\right) = (2 + 3) + 4\left(25 + \frac{25}{4}\right).$$

Inside the brackets:
$$25 + \frac{25}{4} = \frac{100}{4} + \frac{25}{4} = \frac{125}{4}.$$ Multiplying by 4 gives $$125$$, so

$$(\alpha + \beta) + 4(r_1^{2} + r_2^{2}) = 5 + 125 = 130.$$

The value is 130, which matches Option B.

The circle equation is $$x^2 + (y-1)^2 = 1$$, which expands to $$x^2 + y^2 - 2y = 0$$. The origin $$(0,0)$$ lies on this circle since $$0^2 + (0-1)^2 = 1$$.

Consider a chord drawn from the origin to another point $$(x_1, y_1)$$ on the circle. Let the midpoint of this chord be $$M(h, k)$$. Since $$M$$ is the midpoint, $$h = \frac{0 + x_1}{2} = \frac{x_1}{2}$$ and $$k = \frac{0 + y_1}{2} = \frac{y_1}{2}$$, so $$x_1 = 2h$$ and $$y_1 = 2k$$.

Substitute $$(x_1, y_1)$$ into the circle equation: $$(2h)^2 + (2k - 1)^2 = 1$$ $$4h^2 + 4k^2 - 4k + 1 = 1$$ $$4h^2 + 4k^2 - 4k = 0$$ Divide by 4: $$h^2 + k^2 - k = 0$$

Thus, the locus of the midpoints is $$x^2 + y^2 - y = 0$$. Completing the square: $$x^2 + y^2 - y = 0$$ $$x^2 + \left(y^2 - y + \frac{1}{4}\right) = \frac{1}{4}$$ $$x^2 + \left(y - \frac{1}{2}\right)^2 = \left(\frac{1}{2}\right)^2$$ This is a circle with center $$\left(0, \frac{1}{2}\right)$$ and radius $$\frac{1}{2}$$.

This locus circle intersects the line $$x + y = 1$$. Substitute $$y = 1 - x$$ into the locus equation: $$x^2 + \left((1 - x) - \frac{1}{2}\right)^2 = \frac{1}{4}$$ $$x^2 + \left(\frac{1}{2} - x\right)^2 = \frac{1}{4}$$ Expand $$\left(\frac{1}{2} - x\right)^2 = \frac{1}{4} - x + x^2$$: $$x^2 + \frac{1}{4} - x + x^2 = \frac{1}{4}$$ $$2x^2 - x + \frac{1}{4} = \frac{1}{4}$$ $$2x^2 - x = 0$$ $$x(2x - 1) = 0$$ So $$x = 0$$ or $$x = \frac{1}{2}$$.

Corresponding $$y$$-values: - If $$x = 0$$, $$y = 1 - 0 = 1$$, so point $$P(0, 1)$$. - If $$x = \frac{1}{2}$$, $$y = 1 - \frac{1}{2} = \frac{1}{2}$$, so point $$Q\left(\frac{1}{2}, \frac{1}{2}\right)$$.

The distance $$PQ$$ is: $$\sqrt{\left(\frac{1}{2} - 0\right)^2 + \left(\frac{1}{2} - 1\right)^2} = \sqrt{\left(\frac{1}{2}\right)^2 + \left(-\frac{1}{2}\right)^2} = \sqrt{\frac{1}{4} + \frac{1}{4}} = \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}}$$

Therefore, the length of $$PQ$$ is $$\frac{1}{\sqrt{2}}$$, corresponding to option A.

The given circle is $$x^2 + y^2 - 10x - 6y + 30 = 0$$.
To locate its centre and radius we complete the squares.

$$(x^2 - 10x) + (y^2 - 6y) + 30 = 0$$
Add and subtract $$25$$ in the first bracket and $$9$$ in the second bracket:

$$(x^2 - 10x + 25) + (y^2 - 6y + 9) + 30 - 25 - 9 = 0$$

$$(x - 5)^2 + (y - 3)^2 - 4 = 0$$

Hence the centre is $$C(5,\,3)$$ and the radius is $$r = 2$$.

For a square inscribed in a circle, the diagonals are diameters of the circle. Therefore the point of intersection of the diagonals is the centre of the circle, i.e. $$C(5,3)$$ is also the centre of the square.

If the side length of the square is $$s$$ and its circum-radius is $$R$$, the relation is $$R = \dfrac{\text{diagonal}}{2} = \dfrac{s\sqrt{2}}{2} = \dfrac{s}{\sqrt{2}}$$ $$-(1)$$

Putting $$R = r = 2$$ in $$(1)$$: $$s = R\sqrt{2} = 2\sqrt{2}$$.

One side of the square is parallel to $$y = x + 3$$, whose slope is $$1$$. A square whose sides have slopes $$\pm 1$$ has its diagonals along the coordinate axes. Thus, starting from the centre $$C(5,3)$$, the vertices are reached by moving a distance $$R = 2$$ along the positive and negative coordinate axes.

Therefore the four vertices are $$P_1(5+2,\,3) = (7,\,3),\; P_2(5-2,\,3) = (3,\,3),$$ $$P_3(5,\,3+2) = (5,\,5),\; P_4(5,\,3-2) = (5,\,1).$$

Now compute $$x_i^2 + y_i^2$$ for each vertex:

For $$P_1(7,3):\; 7^2 + 3^2 = 49 + 9 = 58$$

For $$P_2(3,3):\; 3^2 + 3^2 = 9 + 9 = 18$$

For $$P_3(5,5):\; 5^2 + 5^2 = 25 + 25 = 50$$

For $$P_4(5,1):\; 5^2 + 1^2 = 25 + 1 = 26$$

The required sum is $$\sum (x_i^2 + y_i^2) = 58 + 18 + 50 + 26 = 152.$$

Hence the correct option is Option B: $$152$$.

$$\frac{(x-2)^2+(y-1)^2}{(x-1)^2+(y-3)^2} = 25/16$$.

$$16[(x-2)^2+(y-1)^2] = 25[(x-1)^2+(y-3)^2]$$.

$$16(x²-4x+4+y²-2y+1) = 25(x²-2x+1+y²-6y+9)$$.

$$16x²-64x+80+16y²-32y = 25x²-50x+250+25y²-150y$$.

$$9x²+9y²+14x-118y+170 = 0$$. So a=9, b=9, c=0, d=14, e=-118.

a²+2b+3c+4d+e = 81+18+0+56-118 = 37.

The correct answer is Option (1): 37.

By the symmetry of the parabola $$y = 6 - x^2$$ and the lines $$y = \sqrt{3}|x|$$ about the $$y$$-axis, the circle of minimum area touching both must be centred on the $$y$$-axis at $$(0, k)$$ with radius $$r$$.

The perpendicular distance from $$(0, k)$$ to the line $$\sqrt{3}\,x - y = 0$$ is $$\dfrac{k}{\sqrt{3+1}} = \dfrac{k}{2}$$. For the circle to touch this line, we need $$r = k/2$$, so $$k = 2r$$.

The lines $$y = \sqrt{3}|x|$$ and the parabola $$y = 6 - x^2$$ intersect where $$6 - x^2 = \sqrt{3}\,x$$, giving $$x^2 + \sqrt{3}\,x - 6 = 0$$, so $$x = \sqrt{3}$$ (positive root) and the intersection point is $$(\sqrt{3}, 3)$$.

For the minimum-area circle, we require it to pass through these intersection points $$(\pm\sqrt{3}, 3)$$. Substituting $$(\sqrt{3}, 3)$$ into the circle equation $$x^2 + (y - 2r)^2 = r^2$$: $$3 + (3 - 2r)^2 = r^2$$, which gives $$3 + 9 - 12r + 4r^2 = r^2$$, i.e., $$3r^2 - 12r + 12 = 0$$, so $$r^2 - 4r + 4 = 0$$, yielding $$(r - 2)^2 = 0$$, hence $$r = 2$$.

The circle has centre $$(0, 4)$$ and radius $$2$$: $$x^2 + (y - 4)^2 = 4$$. Checking option (D), the point $$(2, 4)$$: $$4 + 0 = 4$$ ✓. The answer is $$\boxed{(2, 4)}$$, i.e., Option (D).

Given $$C_1: x^2 + y^2 - 2(x+y) + 1 = 0$$ and $$C_2$$ has centre $$(-1, 0)$$ and radius 2.

We rewrite $$C_1$$ as $$ x^2 - 2x + y^2 - 2y + 1 = 0 $$ which becomes $$ (x-1)^2 - 1 + (y-1)^2 - 1 + 1 = 0 $$ and hence $$ (x-1)^2 + (y-1)^2 = 1. $$

The centre of $$C_1$$ is $$(1, 1)$$ and the radius is 1.

For $$C_2$$ with centre $$(-1,0)$$ and radius 2 the equation is $$ (x+1)^2 + y^2 = 4, $$ which expands to $$ x^2 + 2x + 1 + y^2 = 4 $$ and simplifies to $$ x^2 + y^2 + 2x - 3 = 0. $$

The common chord is obtained by subtracting the equations of $$C_2$$ from $$C_1$$: $$ (x^2 + y^2 - 2x - 2y + 1) - (x^2 + y^2 + 2x - 3) = 0, $$ giving $$ -2x - 2y + 1 - 2x + 3 = 0, $$ or $$ -4x - 2y + 4 = 0 $$ which simplifies to $$ 2x + y - 2 = 0. $$

Setting $$x=0$$ in the chord equation yields $$2(0)+y-2=0$$, so $$y=2$$, and hence $$P=(0,2)$$.

The square of the distance from $$P$$ to the centre of $$C_1$$ is $$ d^2 = (0-1)^2 + (2-1)^2 = 1 + 1 = 2. $$

The correct answer is Option (1): 2.

Let the required circle have centre $$(h,k)$$ and radius $$R$$.

The circle passes through $$(0,0)$$ and $$(1,0)$$, so

$$h^{2}+k^{2}=R^{2} \quad -(1)$$
$$(h-1)^{2}+k^{2}=R^{2} \quad -(2)$$

Subtract $$(1)$$ from $$(2)$$:

$$(h-1)^{2}-h^{2}=0 \;\Longrightarrow\; h^{2}-2h+1-h^{2}=0 \;\Longrightarrow\; -2h+1=0$$

Hence $$h=\frac12$$.

The given circle $$x^{2}+y^{2}=9$$ has centre $$(0,0)$$ and radius $$3$$. Distance between the two centres is therefore

$$D=\sqrt{h^{2}+k^{2}}=R \quad\text{(from }(1)\text{)}.$$

For two circles to be tangent internally, the distance between their centres equals the difference of their radii: $$|3-R|=D$$.

Substituting $$D=R$$ gives $$|3-R|=R$$, which splits into

$$3-R=R \;\Longrightarrow\; 3=2R \;\Longrightarrow\; R=\frac32$$ (possible because $$3\gt R$$),
or $$R-3=R \;\Longrightarrow\; 3=0$$ (impossible).

Thus $$R=\frac32$$, and from $$(1)$$

$$h^{2}+k^{2}=R^{2}=\left(\frac32\right)^{2}=\frac94.$$(Since $$h=\frac12$$)

$$\left(\frac12\right)^{2}+k^{2}=\frac94 \;\Longrightarrow\; \frac14+k^{2}=\frac94 \;\Longrightarrow\; k^{2}=2 \;\Longrightarrow\; k=\pm\sqrt2.$$

Finally,

$$4(h^{2}+k^{2})=4\left(\frac94\right)=9.$$

Therefore, for both possible centres $$\left(\frac12,\sqrt2\right)$$ and $$\left(\frac12,-\sqrt2\right)$$, the value of $$4(h^{2}+k^{2})$$ is $$\mathbf{9}$$.

Circle $$(x-\alpha)^2 + (y-\beta)^2 = 50$$ with $$\alpha, \beta > 0$$ touches $$y + x = 0$$.

Distance from center $$(\alpha, \beta)$$ to the line $$x + y = 0$$:

$$\frac{|\alpha + \beta|}{\sqrt{2}} = \sqrt{50} = 5\sqrt{2}$$

Since $$\alpha, \beta > 0$$: $$\alpha + \beta = 10$$.

Point of tangency P has distance $$4\sqrt{2}$$ from origin.

P lies on $$x + y = 0$$, so $$P = (t, -t)$$ with $$|P| = |t|\sqrt{2} = 4\sqrt{2}$$, so $$|t| = 4$$, $$P = (4, -4)$$ or $$(-4, 4)$$.

P must be the foot of perpendicular from center to the line. The line from $$(\alpha, \beta)$$ perpendicular to $$x + y = 0$$ has direction $$(1, 1)$$.

$$P = (\alpha, \beta) - \frac{\alpha+\beta}{\sqrt{2}} \cdot \frac{(1,1)}{\sqrt{2}} = (\alpha - 5, \beta - 5)$$

If $$P = (4, -4)$$: $$\alpha = 9, \beta = 1$$. $$\alpha + \beta = 10$$ ✓.

If $$P = (-4, 4)$$: $$\alpha = 1, \beta = 9$$. $$\alpha + \beta = 10$$ ✓.

$$(\alpha + \beta)^2 = 100$$.

The answer is $$\boxed{100}$$.

The orthocentre of the triangle formed by the lines $$2x + 3y - 1 = 0$$, $$x + 2y - 1 = 0$$, and $$ax + by - 1 = 0$$ is given to be the centroid of another triangle. For this other triangle, the circumcentre is $$(3, 4)$$ and the orthocentre is $$(-6, -8)$$.

First, recall that in any triangle, the centroid $$G$$, circumcentre $$O$$, and orthocentre $$H$$ lie on the Euler line, and the centroid divides the segment joining the orthocentre and circumcentre in the ratio $$2:1$$, with the circumcentre closer to the centroid. The position vector of the centroid is given by:

$$\vec{G} = \frac{\vec{H} + 2\vec{O}}{3}$$

Substituting the given points $$O' = (3, 4)$$ and $$H' = (-6, -8)$$:

$$G' = \left( \frac{-6 + 2 \times 3}{3}, \frac{-8 + 2 \times 4}{3} \right) = \left( \frac{-6 + 6}{3}, \frac{-8 + 8}{3} \right) = \left( \frac{0}{3}, \frac{0}{3} \right) = (0, 0)$$

Thus, the centroid $$G'$$ is $$(0, 0)$$. This centroid is also the orthocentre of the triangle formed by the lines $$2x + 3y - 1 = 0$$, $$x + 2y - 1 = 0$$, and $$ax + by - 1 = 0$$. Therefore, the orthocentre of this triangle is $$(0, 0)$$.

To find the orthocentre of the triangle formed by the three lines, denote the lines as:

L1: $$2x + 3y = 1$$

L2: $$x + 2y = 1$$

L3: $$ax + by = 1$$

The orthocentre $$(0, 0)$$ satisfies the condition that the line from the orthocentre to each vertex is perpendicular to the opposite side. The vertices of the triangle are the pairwise intersections of these lines.

Find the vertices:

1. Intersection of L1 and L2 (vertex A):

Solve $$2x + 3y = 1$$ and $$x + 2y = 1$$.

From L2: $$x = 1 - 2y$$

Substitute into L1: $$2(1 - 2y) + 3y = 1 \implies 2 - 4y + 3y = 1 \implies 2 - y = 1 \implies y = 1$$

Then $$x = 1 - 2(1) = -1$$

So, vertex A: $$(-1, 1)$$

2. Intersection of L1 and L3 (vertex B):

Solve $$2x + 3y = 1$$ and $$ax + by = 1$$.

Using elimination:

Multiply first equation by $$b$$: $$2b x + 3b y = b$$

Multiply second equation by $$3$$: $$3a x + 3b y = 3$$

Subtract: $$(2b x + 3b y) - (3a x + 3b y) = b - 3 \implies (2b - 3a)x = b - 3$$

So, $$x = \frac{b - 3}{2b - 3a}$$

From first equation multiplied by $$a$$: $$2a x + 3a y = a$$

Second equation multiplied by $$2$$: $$2a x + 2b y = 2$$

Subtract: $$(2a x + 3a y) - (2a x + 2b y) = a - 2 \implies (3a - 2b)y = a - 2$$

So, $$y = \frac{a - 2}{3a - 2b} = -\frac{a - 2}{2b - 3a}$$ (since $$3a - 2b = -(2b - 3a)$$)

Thus, vertex B: $$\left( \frac{b - 3}{2b - 3a}, -\frac{a - 2}{2b - 3a} \right)$$

3. Intersection of L2 and L3 (vertex C):

Solve $$x + 2y = 1$$ and $$ax + by = 1$$.

From L2: $$x = 1 - 2y$$

Substitute into L3: $$a(1 - 2y) + b y = 1 \implies a - 2a y + b y = 1 \implies (b - 2a)y = 1 - a$$

So, $$y = \frac{1 - a}{b - 2a}$$

Then $$x = 1 - 2y = 1 - 2\frac{1 - a}{b - 2a} = \frac{b - 2a - 2(1 - a)}{b - 2a} = \frac{b - 2a - 2 + 2a}{b - 2a} = \frac{b - 2}{b - 2a}$$

Thus, vertex C: $$\left( \frac{b - 2}{b - 2a}, \frac{1 - a}{b - 2a} \right)$$

Since the orthocentre is $$(0, 0)$$, the vector from $$(0, 0)$$ to each vertex is perpendicular to the direction vector of the opposite side.

For vertex A$$(-1, 1)$$, the opposite side is BC. Since B and C lie on L3, BC is the line L3: $$ax + by = 1$$. The normal vector to L3 is $$(a, b)$$, so the direction vector of BC is $$(-b, a)$$. The vector OA is $$(-1, 1)$$. Since AH is perpendicular to BC:

$$(-1, 1) \cdot (-b, a) = 0 \implies (-1)(-b) + (1)(a) = b + a = 0 \implies a + b = 0 \quad \text{(Equation 1)}$$

For vertex B$$\left( \frac{b - 3}{2b - 3a}, -\frac{a - 2}{2b - 3a} \right)$$, the opposite side is AC. Since A and C lie on L2, AC is the line L2: $$x + 2y = 1$$. The normal vector to L2 is $$(1, 2)$$, so the direction vector of AC is $$(-2, 1)$$. The vector OB is $$\left( \frac{b - 3}{2b - 3a}, -\frac{a - 2}{2b - 3a} \right)$$. Since BH is perpendicular to AC:

$$\left( \frac{b - 3}{2b - 3a} \right)(-2) + \left( -\frac{a - 2}{2b - 3a} \right)(1) = 0$$

Since the denominator $$2b - 3a \neq 0$$ (assuming a non-degenerate triangle), multiply by $$2b - 3a$$:

$$-2(b - 3) - (a - 2) = 0 \implies -2b + 6 - a + 2 = 0 \implies -a - 2b + 8 = 0 \implies a + 2b = 8 \quad \text{(Equation 2)}$$

Solving Equations 1 and 2:

From Equation 1: $$a + b = 0 \implies a = -b$$

Substitute into Equation 2: $$-b + 2b = 8 \implies b = 8$$

Then $$a = -b = -8$$

So, $$a = -8$$, $$b = 8$$

Now, verify for vertex C$$\left( \frac{b - 2}{b - 2a}, \frac{1 - a}{b - 2a} \right)$$. With $$a = -8$$, $$b = 8$$:

$$x_c = \frac{8 - 2}{8 - 2(-8)} = \frac{6}{8 + 16} = \frac{6}{24} = \frac{1}{4}$$

$$y_c = \frac{1 - (-8)}{8 - 2(-8)} = \frac{9}{24} = \frac{3}{8}$$

So, vertex C: $$\left( \frac{1}{4}, \frac{3}{8} \right)$$

The opposite side is AB. Since A and B lie on L1, AB is the line L1: $$2x + 3y = 1$$. The normal vector to L1 is $$(2, 3)$$, so the direction vector of AB is $$(-3, 2)$$. The vector OC is $$\left( \frac{1}{4}, \frac{3}{8} \right)$$. Since CH is perpendicular to AB:

$$\left( \frac{1}{4} \right)(-3) + \left( \frac{3}{8} \right)(2) = -\frac{3}{4} + \frac{6}{8} = -\frac{3}{4} + \frac{3}{4} = 0$$

Thus, the condition is satisfied.

Check the denominators for non-zero values:

$$2b - 3a = 2(8) - 3(-8) = 16 + 24 = 40 \neq 0$$

$$b - 2a = 8 - 2(-8) = 8 + 16 = 24 \neq 0$$

The slopes of the lines are distinct: L1 has slope $$-\frac{2}{3}$$, L2 has slope $$-\frac{1}{2}$$, L3 (with $$a = -8$$, $$b = 8$$) is $$-8x + 8y = 1$$, so slope $$\frac{8}{8} = 1$$. Thus, no parallel lines, and a triangle is formed.

Now, compute $$|a - b|$$:

$$|a - b| = |-8 - 8| = |-16| = 16$$

Therefore, the value of $$|a - b|$$ is 16.

$$C_1: x^2+y^2=25$$ (center O₁(0,0), r₁=5). $$C_2: (x-\alpha)^2+y^2=16$$ (center O₂(α,0), r₂=4).

The angle between radii at intersection point is $$\sin^{-1}\frac{\sqrt{63}}{8}$$.

Let the angle be $$\theta$$ at the intersection point P. Using the law of cosines in triangle O₁PO₂:

$$\alpha^2 = 25 + 16 - 2(5)(4)\cos\theta = 41 - 40\cos\theta$$

where $$\theta$$ is the angle at P. $$\sin\theta = \frac{\sqrt{63}}{8}$$, $$\cos\theta = \pm\frac{1}{8}$$.

Since $$\alpha \in (5,9)$$, $$\alpha^2 \in (25,81)$$. With $$\cos\theta = 1/8$$: $$\alpha^2 = 41-5 = 36$$, $$\alpha = 6$$ ✓.

Common chord length: The perpendicular from O₁ to the radical axis. Radical axis: subtract circles: $$-2\alpha x + \alpha^2 = -9 \Rightarrow x = \frac{\alpha^2+9}{2\alpha} = \frac{45}{12} = \frac{15}{4}$$.

$$\beta = 2\sqrt{r_1^2 - x^2} = 2\sqrt{25 - 225/16} = 2\sqrt{175/16} = 2 \cdot \frac{5\sqrt{7}}{4} = \frac{5\sqrt{7}}{2}$$.

$$(\alpha\beta)^2 = (6 \cdot \frac{5\sqrt{7}}{2})^2 = (15\sqrt{7})^2 = 225 \times 7 = 1575$$.

Therefore, the answer is $$\boxed{1575}$$.

The circle is given by $$x^2 + y^2 = 169$$, which has center $$(0,0)$$ and radius $$13$$. The line is $$5x - y = 13$$. The area below this line and inside the circle needs to be found.

First, find the points of intersection between the circle and the line by substituting $$y = 5x - 13$$ into the circle equation:

$$x^2 + (5x - 13)^2 = 169$$

$$x^2 + 25x^2 - 130x + 169 = 169$$

$$26x^2 - 130x = 0$$

$$2x(13x - 65) = 0$$

So, $$x = 0$$ or $$x = 5$$.

When $$x = 0$$, $$y = 5(0) - 13 = -13$$.

When $$x = 5$$, $$y = 5(5) - 13 = 12$$.

The points of intersection are $$(0, -13)$$ and $$(5, 12)$$.

The chord joining these points subtends an angle $$\theta$$ at the center. The vectors are $$\overrightarrow{OA} = (0, -13)$$ and $$\overrightarrow{OB} = (5, 12)$$. The dot product is:

$$\overrightarrow{OA} \cdot \overrightarrow{OB} = (0)(5) + (-13)(12) = -156$$

$$|\overrightarrow{OA}| = 13, \quad |\overrightarrow{OB}| = 13$$

$$\cos \theta = \frac{-156}{13 \times 13} = -\frac{12}{13}$$

Thus, $$\theta = \cos^{-1}\left(-\frac{12}{13}\right)$$. Since $$\cos \theta = -\frac{12}{13}$$, $$\theta$$ is obtuse, and $$\sin \theta = \sqrt{1 - \left(-\frac{12}{13}\right)^2} = \sqrt{\frac{25}{169}} = \frac{5}{13}$$ (positive as $$\theta$$ is in the second quadrant).

Let $$\gamma = \cos^{-1}\left(\frac{12}{13}\right)$$, so $$\theta = \pi - \gamma$$ and $$\sin \gamma = \frac{5}{13}$$.

The area of the minor segment (below the chord) is given by the formula $$\frac{r^2}{2} (\theta - \sin \theta)$$:

$$\text{Area} = \frac{169}{2} \left( \theta - \frac{5}{13} \right)$$

Substitute $$\theta = \pi - \gamma$$:

$$\text{Area} = \frac{169}{2} \left( (\pi - \gamma) - \frac{5}{13} \right) = \frac{169}{2} (\pi - \gamma) - \frac{169}{2} \cdot \frac{5}{13}$$

$$\frac{169}{2} \cdot \frac{5}{13} = \frac{169 \times 5}{2 \times 13} = \frac{13 \times 13 \times 5}{2 \times 13} = \frac{65}{2}$$

So,

$$\text{Area} = \frac{169}{2} (\pi - \gamma) - \frac{65}{2}$$

Now, $$\gamma = \cos^{-1}\left(\frac{12}{13}\right) = \sin^{-1}\left(\frac{5}{13}\right)$$. Let $$\beta = \sin^{-1}\left(\frac{12}{13}\right)$$, so $$\sin \beta = \frac{12}{13}$$ and $$\cos \beta = \frac{5}{13}$$. Then $$\gamma = \frac{\pi}{2} - \beta$$, since $$\sin(\gamma + \beta) = \sin \gamma \cos \beta + \cos \gamma \sin \beta = \left(\frac{5}{13}\right)\left(\frac{5}{13}\right) + \left(\frac{12}{13}\right)\left(\frac{12}{13}\right) = 1$$, and $$\gamma + \beta = \frac{\pi}{2}$$.

Substitute $$\gamma = \frac{\pi}{2} - \beta$$:

$$\text{Area} = \frac{169}{2} \left( \pi - \left(\frac{\pi}{2} - \beta\right) \right) - \frac{65}{2} = \frac{169}{2} \left( \pi - \frac{\pi}{2} + \beta \right) - \frac{65}{2}$$

$$= \frac{169}{2} \left( \frac{\pi}{2} + \beta \right) - \frac{65}{2} = \frac{169}{2} \cdot \frac{\pi}{2} + \frac{169}{2} \beta - \frac{65}{2} = \frac{169\pi}{4} + \frac{169}{2} \beta - \frac{65}{2}$$

Since $$\beta = \sin^{-1}\left(\frac{12}{13}\right)$$,

$$\text{Area} = \frac{169\pi}{4} + \frac{169}{2} \sin^{-1}\left(\frac{12}{13}\right) - \frac{65}{2}$$

The given expression is $$\frac{\pi \alpha}{2\beta} - \frac{65}{2} + \frac{\alpha}{\beta} \sin^{-1}\left(\frac{12}{13}\right)$$, where $$\alpha$$ and $$\beta$$ are coprime integers. Comparing coefficients:

Coefficient of $$\pi$$: $$\frac{\alpha}{2\beta} = \frac{169}{4}$$

Coefficient of $$\sin^{-1}\left(\frac{12}{13}\right)$$: $$\frac{\alpha}{\beta} = \frac{169}{2}$$

Constant term: $$-\frac{65}{2}$$ matches.

From $$\frac{\alpha}{\beta} = \frac{169}{2}$$, and $$\frac{\alpha}{2\beta} = \frac{1}{2} \cdot \frac{\alpha}{\beta} = \frac{1}{2} \cdot \frac{169}{2} = \frac{169}{4}$$, which matches the coefficient of $$\pi$$. Thus, $$\frac{\alpha}{\beta} = \frac{169}{2}$$, so $$\alpha = 169$$, $$\beta = 2$$. These are coprime since $$169 = 13^2$$ and $$2$$ have no common factors.

Therefore, $$\alpha + \beta = 169 + 2 = 171$$.

The equation $$\left|\frac{z-2}{z-3}\right| = 2$$ means $$|z - 2| = 2|z - 3|$$.

Let $$z = x + iy$$. Squaring both sides:

$$(x - 2)^2 + y^2 = 4\left[(x - 3)^2 + y^2\right]$$

Expanding: $$x^2 - 4x + 4 + y^2 = 4x^2 - 24x + 36 + 4y^2$$

Simplifying: $$3x^2 + 3y^2 - 20x + 32 = 0$$

Dividing by 3: $$x^2 + y^2 - \frac{20}{3}x + \frac{32}{3} = 0$$

Completing the square for $$x$$:

$$\left(x - \frac{10}{3}\right)^2 + y^2 = \frac{100}{9} - \frac{32}{3} = \frac{100 - 96}{9} = \frac{4}{9}$$

This is a circle with center $$\left(\frac{10}{3},\, 0\right)$$ and radius $$\frac{2}{3}$$.

So $$\alpha = \frac{10}{3}$$, $$\beta = 0$$, and $$\gamma = \frac{2}{3}$$.

$$\alpha + \beta + \gamma = \frac{10}{3} + 0 + \frac{2}{3} = \frac{12}{3} = 4$$

Therefore $$3(\alpha + \beta + \gamma) = 3 \times 4 = 12$$

The answer is Option D: $$12$$.

The circle $$x^2 + y^2 - 2x = 0$$ can be written as $$(x-1)^2 + y^2 = 1$$, with center $$(1, 0)$$ and radius $$1$$.

Point $$A(\alpha,0)$$ lies on the circle, so $$\alpha^2 - 2\alpha = 0$$ and thus $$\alpha(\alpha - 2) = 0$$ gives $$\alpha = 0$$ or $$\alpha = 2$$. Since $$A$$ also lies on $$ax + by = 0$$, we have $$a\alpha = 0$$. If $$a = 0$$ then the line is $$y = 0$$, intersecting the circle at $$(0,0)$$ and $$(2,0)$$, but then a point $$B(1,\beta)$$ with $$\beta \neq 0$$ could not lie on this line. Therefore $$a \neq 0$$, which forces $$\alpha = 0$$ and gives $$A = (0,0)$$.

Point $$B(1,\beta)$$ lies on the circle, so $$1 + \beta^2 - 2 = 0$$ and hence $$\beta^2 = 1$$ leading to $$\beta = \pm 1$$. It also lies on $$ax + by = 0$$, so $$a + b\beta = 0$$ and $$\beta = -a/b$$. If $$\beta = 1$$ then $$a = -b$$, which is allowed since $$a \neq b$$; if $$\beta = -1$$ then $$a = b$$, contradicting $$a \neq b$$. Thus $$\beta = 1$$ and $$B = (1,1)$$.

With $$A = (0,0)$$ and $$B = (1,1)$$, the midpoint of $$AB$$ is $$\left(\frac{0+1}{2}, \frac{0+1}{2}\right) = \left(\frac{1}{2}, \frac{1}{2}\right)$$ and the square of the radius of the circle with $$AB$$ as diameter is $$\left(\frac{1}{2}\right)^2 + \left(\frac{1}{2}\right)^2 = \frac{1}{2}$$.

To reflect the center across the line $$x + y + 2 = 0$$, we use the formula for reflecting $$(x_0,y_0)$$ across $$ax + by + c = 0$$:
$$x' = x_0 - \frac{2a(ax_0 + by_0 + c)}{a^2 + b^2}, \quad y' = y_0 - \frac{2b(ax_0 + by_0 + c)}{a^2 + b^2}.$$
Here $$a = 1$$, $$b = 1$$, $$c = 2$$, and $$(x_0,y_0) = \left(\frac{1}{2}, \frac{1}{2}\right)$$. Then
$$ax_0 + by_0 + c = \frac{1}{2} + \frac{1}{2} + 2 = 3,$$
$$x' = \frac{1}{2} - \frac{2(1)(3)}{2} = \frac{1}{2} - 3 = -\frac{5}{2},$$
$$y' = \frac{1}{2} - \frac{2(1)(3)}{2} = \frac{1}{2} - 3 = -\frac{5}{2}.$$
Thus the reflected center is $$\left(-\frac{5}{2}, -\frac{5}{2}\right)$$.

Reflection preserves distances, so the radius squared of the image circle remains $$\frac{1}{2}$$. Its equation is
$$\left(x + \frac{5}{2}\right)^2 + \left(y + \frac{5}{2}\right)^2 = \frac{1}{2},$$
which expands to
$$x^2 + 5x + \frac{25}{4} + y^2 + 5y + \frac{25}{4} = \frac{1}{2},$$
then to
$$x^2 + y^2 + 5x + 5y + \frac{50}{4} - \frac{2}{4} = 0,$$
and finally to
$$x^2 + y^2 + 5x + 5y + 12 = 0.$$
Therefore the correct answer is Option A: $$x^2 + y^2 + 5x + 5y + 12 = 0.$$

Find the area of trapezium AMNB where A, B are centers of circles $$C_1$$, $$C_2$$ and M, N are their feet of perpendiculars on the x-axis.

The given circle can be written as $$x^2 + y^2 - 4x - 6y + 11 = 0 \Rightarrow (x-2)^2 + (y-3)^2 = 2$$, which shows that its center is $$(2, 3)$$ and its radius is $$\sqrt{2}$$.

The tangent at the point $$(3, 2)$$ on this circle is obtained from the formula $$(x-2)(3-2) + (y-3)(2-3) = 2$$, which simplifies to $$(x-2) - (y-3) = 2$$ or $$x - y = 1$$.

When the circle rolls 4 units upward along this tangent line, its center moves in the direction of the unit vector along $$x - y = 1$$, namely $$\left(\tfrac{1}{\sqrt{2}}, \tfrac{1}{\sqrt{2}}\right)$$. Hence the new center A of circle $$C_1$$ is $$A = \left(2 + \frac{4}{\sqrt{2}},\,3 + \frac{4}{\sqrt{2}}\right) = (2 + 2\sqrt{2},\,3 + 2\sqrt{2}).$$

The image of this center in the line $$x - y - 1 = 0$$ gives the center B of circle $$C_2$$. Reflecting $$(a, b)$$ in this line yields $$x' = b + 1,\quad y' = a - 1,$$ so $$B = (3 + 2\sqrt{2} + 1,\;2 + 2\sqrt{2} - 1) = (4 + 2\sqrt{2},\,1 + 2\sqrt{2}).$$

The feet of the perpendiculars from A and B onto the x-axis are $$M = (2 + 2\sqrt{2},\,0)$$ and $$N = (4 + 2\sqrt{2},\,0),$$ respectively.

The heights of A and B above the x-axis are $$AM = 3 + 2\sqrt{2},\quad BN = 1 + 2\sqrt{2},$$ and the distance between M and N is $$MN = (4 + 2\sqrt{2}) - (2 + 2\sqrt{2}) = 2.$$ Therefore the area of trapezium AMNB is $$\frac{1}{2}(AM + BN)\times MN = \frac{1}{2}(3 + 2\sqrt{2} + 1 + 2\sqrt{2})\times 2 = 4 + 4\sqrt{2} = 4(1 + \sqrt{2}).$$

The correct answer is Option B: $$4(1 + \sqrt{2})$$.

A chord AB of length $$\lambda$$ moves on a circle of radius $$\lambda$$, and a point P divides AB in the ratio 2:3. We wish to find the locus of P.

To begin, take the circle to be $$x^2 + y^2 = \lambda^2$$ with endpoints of the chord given by $$A=(\lambda\cos\alpha,\;\lambda\sin\alpha)$$ and $$B=(\lambda\cos\beta,\;\lambda\sin\beta)\,.$$

Imposing the condition that the length of the chord $$AB$$ equals $$\lambda$$ leads to $$ |AB|^2 = 2\lambda^2 - 2\lambda^2\cos(\alpha-\beta) = \lambda^2 $$ and hence $$ \cos(\alpha-\beta)=\tfrac12\implies\alpha-\beta=\pm\tfrac{\pi}{3}\,. $$

Since P divides AB in the ratio 2:3, its coordinates are $$ P=\frac{3A+2B}{5}=\frac{\lambda}{5}(3\cos\alpha+2\cos\beta,\;3\sin\alpha+2\sin\beta)\,. $$ Writing $$h=\frac{\lambda}{5}(3\cos\alpha+2\cos\beta)$$ and $$k=\frac{\lambda}{5}(3\sin\alpha+2\sin\beta)$$ gives the coordinates of P as $$(h,k)\,.$$

Next, compute $$ h^2+k^2=\frac{\lambda^2}{25}\bigl[(3\cos\alpha+2\cos\beta)^2+(3\sin\alpha+2\sin\beta)^2\bigr] $$ which simplifies to $$ \frac{\lambda^2}{25}\bigl[9+12\cos(\alpha-\beta)+4\bigr]=\frac{\lambda^2}{25}\bigl[13+12\times\tfrac12\bigr]=\frac{19\lambda^2}{25}\,. $$

Therefore, the locus of P is the circle $$x^2+y^2=\tfrac{19\lambda^2}{25}$$, whose radius is $$\tfrac{\sqrt{19}}{5}\lambda\,$$ and this corresponds to Option 3: $$\frac{\sqrt{19}}{5}\lambda\,. $$

The circle is $$x^2 + y^2 - 2x + y - 5 = 0$$ with centre $$C = \bigl(1, -\frac{1}{2}\bigr)$$ and $$r^2 = 1 + \tfrac{1}{4} + 5 = \tfrac{25}{4}$$, so $$r = \tfrac{5}{2}$$. The tangents from $$R\bigl(\tfrac{9}{4},2\bigr)$$ touch the circle at $$P$$ and $$Q$$. To find the equation of the chord of contact $$PQ$$ we use the formula for the given circle where $$g = -1$$, $$f = \tfrac{1}{2}$$ and $$c = -5$$, namely $$ xx_1 + yy_1 + g(x + x_1) + f(y + y_1) + c = 0. $$ Substituting $$R = \bigl(\tfrac{9}{4},2\bigr)$$ into this equation gives $$ \frac{9x}{4} + 2y - \Bigl(x + \frac{9}{4}\Bigr) + \frac{1}{2}(y + 2) - 5 = 0, $$ which simplifies to $$ \frac{5x}{4} + \frac{5y}{2} - \frac{25}{4} = 0 \;\Longrightarrow\; x + 2y - 5 = 0. $$

Substituting $$x = 5 - 2y$$ into the circle equation yields $$ (5 - 2y)^2 + y^2 - 2(5 - 2y) + y - 5 = 0 $$ or $$ 5y^2 - 15y + 10 = 0 \;\Longrightarrow\; y^2 - 3y + 2 = 0. $$ Hence $$y = 1$$ or $$y = 2$$, which gives the points $$P = (3,1)$$ and $$Q = (1,2)$$.

Using the standard coordinate formula for the area of triangle $$PQR$$ with $$P = (3,1)$$, $$Q = (1,2)$$ and $$R = \bigl(\tfrac{9}{4},2\bigr)$$, we have $$ \text{Area} = \frac{1}{2}\bigl|x_P(y_Q - y_R) + x_Q(y_R - y_P) + x_R(y_P - y_Q)\bigr| = \frac{1}{2}\bigl|3(2 - 2) + 1(2 - 1) + \tfrac{9}{4}(1 - 2)\bigr| = \frac{1}{2}\bigl|0 + 1 - \tfrac{9}{4}\bigr| = \frac{1}{2} \cdot \frac{5}{4} = \frac{5}{8}. $$

The correct answer is Option C: $$\dfrac{5}{8}$$.

We need to find the centre $$C(\alpha,\beta)$$ of the locus of point P that divides AB in the ratio 3:2 where A(1,2) and B lies on $$x^2 + y^2 = 16$$, and then determine the length of the line segment AC.

Since B lies on the circle $$x^2 + y^2 = 16$$ of radius 4 centered at the origin, we may parametrize it as $$B = (4\cos\theta, 4\sin\theta)$$. The point P divides AB in the ratio 3:2 from A to B, so by the section formula
$$P = \frac{2A + 3B}{5} = \left(\frac{2 + 12\cos\theta}{5}, \frac{4 + 12\sin\theta}{5}\right).$$

Writing $$P = (h,k)$$ leads to
$$h = \frac{2 + 12\cos\theta}{5}\;\implies\;\cos\theta = \frac{5h - 2}{12},$$
$$k = \frac{4 + 12\sin\theta}{5}\;\implies\;\sin\theta = \frac{5k - 4}{12}.$$ Using the identity $$\cos^2\theta + \sin^2\theta = 1$$ gives
$$\left(\frac{5h-2}{12}\right)^2 + \left(\frac{5k-4}{12}\right)^2 = 1,$$ which simplifies to
$$(5h-2)^2 + (5k-4)^2 = 144.$$ This is the equation of a circle with centre determined by $$5h - 2 = 0$$ and $$5k - 4 = 0$$, namely
$$h = \frac{2}{5},\quad k = \frac{4}{5}.$$ Hence
$$C = \left(\frac{2}{5}, \frac{4}{5}\right).$$

The distance $$AC$$ is then
$$AC = \sqrt{\left(1 - \frac{2}{5}\right)^2 + \left(2 - \frac{4}{5}\right)^2} = \sqrt{\left(\frac{3}{5}\right)^2 + \left(\frac{6}{5}\right)^2} = \sqrt{\frac{9}{25} + \frac{36}{25}} = \sqrt{\frac{45}{25}} = \sqrt{\frac{9}{5}} = \frac{3}{\sqrt{5}} = \frac{3\sqrt{5}}{5}.$$

The correct answer is Option 1: $$\frac{3\sqrt{5}}{5}$$.

To solve this efficiently, notice the relationship between the slopes of the lines.

• Line 1 ($$L_1$$): $$4x + 3y = 69 \implies m_1 = -4/3$$
• Line 2 ($$L_2$$): $$-3x + 4y = 17 \implies m_2 = 3/4$$
• Since $$m_1 \cdot m_2 = (-4/3) \cdot (3/4) = -1$$, $$L_1$$ and $$L_2$$ are perpendicular. This is a right-angled triangle.

In a right-angled triangle, the circumcentre is the midpoint of the hypotenuse. The hypotenuse is the third line ($$L_3$$): $$x + 7y = 61$$.

Find the vertices by intersecting $$L_3$$ with $$L_1$$ and $$L_2$$:

• Vertex $$A$$ ($$L_1 \cap L_3$$): $$4x + 3y = 69$$ and $$x + 7y = 61$$. Solving gives $$(9, 11)$$.
• Vertex $$B$$ ($$L_2 \cap L_3$$): $$-3x + 4y = 17$$ and $$x + 7y = 61$$. Solving gives $$(5, 8)$$.

Circumcentre $$C(\alpha, \beta)$$ = Midpoint of $$AB$$:

$$\alpha = \frac{9+5}{2} = 7, \quad \beta = \frac{11+8}{2} = 9.5$$

We need $$(\alpha - \beta)^2 + \alpha + \beta$$:

$$(7 - 9.5)^2 + 7 + 9.5$$

$$(-2.5)^2 + 16.5 = 6.25 + 16.5 = \mathbf{22.75}$$

Circle 1: $$x^2 + y^2 - 18x - 15y + 131 = 0$$

Center: $$(9, 7.5)$$, radius: $$r_1 = \sqrt{81 + 56.25 - 131} = \sqrt{6.25} = 2.5$$

Circle 2: $$x^2 + y^2 - 6x - 6y - 7 = 0$$

Center: $$(3, 3)$$, radius: $$r_2 = \sqrt{9 + 9 + 7} = \sqrt{25} = 5$$

Distance between centers:

$$d = \sqrt{(9-3)^2 + (7.5-3)^2} = \sqrt{36 + 20.25} = \sqrt{56.25} = 7.5$$

Now: $$r_1 + r_2 = 2.5 + 5 = 7.5 = d$$

Since $$d = r_1 + r_2$$, the circles are externally tangent.

For externally tangent circles, the number of common tangents = 3 (2 external + 1 internal at the point of tangency).

This matches option 1: 3.

A key property in geometry is that the circumcircle of the triangle formed by an external point and its points of tangency ($$OPQ$$) always has the line segment joining the external point ($$O$$) and the center of the circle ($$C$$) as its diameter.

1. Find the Circle's Center

The given circle is $$x^2 + y^2 - 6x + 4y + 8 = 0$$.

By comparing with $$x^2 + y^2 + 2gx + 2fy + c = 0$$, we get:

• Center $$C = (-g, -f) = (3, -2)$$

2. Equation of the Circumcircle of $$\triangle OPQ$$

Point $$O$$ is $$(0,0)$$ and Point $$C$$ is $$(3,-2)$$. Since $$OC$$ is the diameter, we use the diameter form:

$$(x - x_1)(x - x_2) + (y - y_1)(y - y_2) = 0$$

$$(x - 0)(x - 3) + (y - 0)(y + 2) = 0$$

$$x^2 - 3x + y^2 + 2y = 0$$

3. Solve for $$\alpha$$

The circle passes through $$(\alpha, \frac{1}{2})$$. Substitute these coordinates:

$$\alpha^2 - 3\alpha + \left(\frac{1}{2}\right)^2 + 2\left(\frac{1}{2}\right) = 0$$

$$\alpha^2 - 3\alpha + \frac{1}{4} + 1 = 0$$

$$\alpha^2 - 3\alpha + \frac{5}{4} = 0$$

Multiply by 4:

$$4\alpha^2 - 12\alpha + 5 = 0$$

Factor the quadratic:

$$(2\alpha - 5)(2\alpha - 1) = 0$$

$$\alpha = \frac{5}{2} \text{ or } \alpha = \frac{1}{2}$$

Comparing with the options, the value of $$\alpha$$ is $$\frac{5}{2}$$.

Correct Option: (C)

We need to find $$r_1^2 + r_2^2 - r_1 r_2$$ for two circles in the first quadrant that touch both coordinate axes and each cuts an intercept of 2 units on the line $$x + y = 2$$.

A circle of this type has center $$(r,r)$$ and radius $$r$$.

$$ (x - r)^2 + (y - r)^2 = r^2 $$

In order to determine the chord length cut by the line, first compute the perpendicular distance from the center to that line. For the center $$(r,r)$$ and the line $$x + y - 2 = 0$$, this distance is

$$ d = \frac{|r + r - 2|}{\sqrt{2}} = \frac{|2r - 2|}{\sqrt{2}} $$

Since the intercept (or chord) length on the line is 2 units, the chord length formula $$2\sqrt{r^2 - d^2} = 2$$ implies that

$$ r^2 - d^2 = 1 $$

Substituting the value of $$d^2 = \frac{(2r-2)^2}{2} = \frac{4r^2 - 8r + 4}{2} = 2r^2 - 4r + 2$$ into this equation yields

$$ r^2 - (2r^2 - 4r + 2) = 1 $$

$$ r^2 - 2r^2 + 4r - 2 = 1 $$

$$ -r^2 + 4r - 3 = 0 $$

$$ r^2 - 4r + 3 = 0 $$

Factoring the quadratic expression gives

$$ (r - 1)(r - 3) = 0 $$

Accordingly, the two possible radii are $$r_1 = 1$$ and $$r_2 = 3$$.

Next, computing the desired combination yields

$$ r_1^2 + r_2^2 - r_1 r_2 = 1 + 9 - 3 = 7 $$

Therefore, the required value is $$7$$.

A circle in the first quadrant touches both coordinate axes. It passes through a point $$P(\alpha, \beta)$$ above the line AB, where A and B are the points where the circle touches the x-axis and y-axis respectively. PQ = 11 units, where Q is the foot of perpendicular from P onto line AB.

First, we set up the equation of the circle.

A circle touching both coordinate axes in the first quadrant has its centre at $$(a, a)$$ and radius $$a$$. The equation of the circle is:

$$(x - a)^2 + (y - a)^2 = a^2$$

The circle touches the x-axis at $$A(a, 0)$$ and the y-axis at $$B(0, a)$$.

Next, we write the equation of line AB.

The line joining $$A(a, 0)$$ and $$B(0, a)$$ has the equation:

$$\frac{x}{a} + \frac{y}{a} = 1 \implies x + y = a$$

From this, we use the perpendicular distance formula.

The perpendicular distance from $$P(\alpha, \beta)$$ to the line $$x + y - a = 0$$ is:

$$PQ = \frac{|\alpha + \beta - a|}{\sqrt{1^2 + 1^2}} = \frac{|\alpha + \beta - a|}{\sqrt{2}} = 11$$

Since P is above the line AB, we have $$\alpha + \beta > a$$, so:

$$\alpha + \beta - a = 11\sqrt{2} \implies a = \alpha + \beta - 11\sqrt{2} \quad \cdots (i)$$

Now, we use the condition that P lies on the circle.

Since $$P(\alpha, \beta)$$ lies on the circle:

$$(\alpha - a)^2 + (\beta - a)^2 = a^2$$

Expanding:

$$\alpha^2 - 2a\alpha + a^2 + \beta^2 - 2a\beta + a^2 = a^2$$

$$a^2 - 2a(\alpha + \beta) + \alpha^2 + \beta^2 = 0$$

Then, we rearrange and simplify.

Rewriting: $$(a - (\alpha + \beta))^2 - (\alpha + \beta)^2 + \alpha^2 + \beta^2 = 0$$

$$(a - (\alpha + \beta))^2 = (\alpha + \beta)^2 - \alpha^2 - \beta^2 = 2\alpha\beta$$

Continuing, we substitute from equation (i).

From equation (i), $$a - (\alpha + \beta) = -11\sqrt{2}$$. Substituting:

$$(-11\sqrt{2})^2 = 2\alpha\beta$$

$$242 = 2\alpha\beta$$

$$\alpha\beta = 121$$

The value of $$\alpha\beta$$ is 121.

To solve this, we use the property that a mirror image of a circle has the same radius as the original.

Rewrite $$C_1$$: $$x^2 + y^2 - 4x - 2y + 5 - \alpha = 0$$.

The center is $$(2, 1)$$ and the radius squared ($$r^2$$) is:

$$r^2 = g^2 + f^2 - c = 2^2 + 1^2 - (5 - \alpha) = 5 - 5 + \alpha = /alpha$$

So, $$r = \sqrt{\alpha}$$.

Divide the $$C_2$$ equation by 5: $$x^2 + y^2 - 2fx - 2gy + \frac{36}{5} = 0$$.

The radius squared for $$C_2$$ is:

$$r^2 = f^2 + g^2 - \frac{36}{5}$$

The center of $$C_2$$ $$(f, g)$$ is the mirror image of the center of $$C_1$$ $$(2, 1)$$ in the line $$2x - y + 1 = 0$$.

Using the image formula: $$\frac{f-2}{2} = \frac{g-1}{-1} = -2\frac{2(2) - 1(1) + 1}{2^2 + (-1)^2}$$

$$\frac{f-2}{2} = \frac{g-1}{-1} = -2\frac{4}{5} = -1.6$$

• $$f = 2 + 2(-1.6) = \mathbf{-1.2}$$
• $$g = 1 - 1(-1.6) = \mathbf{2.6}$$

Since radii are equal:

$$\alpha = f^2 + g^2 - \frac{36}{5} = (-1.2)^2 + (2.6)^2 - 7.2$$

$$\alpha = 1.44 + 6.76 - 7.2 = \mathbf{1}$$

Since $$r = \sqrt{\alpha}$$, then $$r = 1$$.

$$\alpha + r = 1 + 1 = \mathbf{2}$$

Points $$P(-3, 2)$$, $$Q(9, 10)$$, and $$R(\alpha, 4)$$ lie on a circle $$C$$ with $$PR$$ as its diameter.

Find the center and radius of circle $$C$$.

Since $$PR$$ is the diameter, the center is the midpoint of $$PR$$:

Use the condition that $$Q(9, 10)$$ lies on the circle.

The radius equals both the distance from center to $$P$$ and to $$Q$$:

Expanding:

So $$R = (13, 4)$$, Center = $$(5, 3)$$, Radius$$^2 = (5+3)^2 + (3-2)^2 = 64 + 1 = 65$$.

Find the tangent at $$Q(9, 10)$$.

Find the tangent at $$R(13, 4)$$.

Find the intersection point $$S$$.

From $$8x + y = 108$$: $$y = 108 - 8x$$

Substituting into $$4x + 7y = 106$$:

So $$S = \left(\frac{25}{2}, 8\right)$$.

Find $$k$$.

$$S$$ lies on $$2x - ky = 1$$:

The value of $$k$$ is $$\boxed{3}$$.

We need to find the circumradius squared of the triangle formed by three lines.

First, consider the tangent to $$y^2 = 2x$$ at $$(2,2)$$. Differentiating $$y^2 = 2x$$ gives $$2y \frac{dy}{dx} = 2$$, so $$\frac{dy}{dx} = \frac{1}{y} = \frac{1}{2}$$ at $$(2,2)$$. Therefore, the tangent line is $$y - 2 = \frac{1}{2}(x - 2)$$, which can be written as $$x - 2y + 2 = 0$$.

Next, consider the tangent to $$x^2 + y^2 = 4x$$ at $$(2,2)$$. Rewriting the equation as $$(x-2)^2 + y^2 = 4$$ shows that this circle has center $$(2,0)$$ and radius 2. The radius drawn to $$(2,2)$$ is vertical, which means the tangent there is horizontal: $$y = 2$$.

The third line is given by $$x + y + 2 = 0$$.

The intersections of these three lines determine the vertices of the triangle. Solving $$x - 2y + 2 = 0$$ with $$y = 2$$ yields $$x = 2$$, so vertex $$A$$ is $$(2,2)$$. Solving $$x - 2y + 2 = 0$$ with $$x + y + 2 = 0$$ (i.e., $$x = -y - 2$$) gives $$y = 0$$ and $$x = -2$$, so vertex $$B$$ is $$(-2,0)$$. Finally, solving $$y = 2$$ with $$x + y + 2 = 0$$ gives $$x = -4$$, so vertex $$C$$ is $$(-4,2)$$.

Let $$a = BC$$, $$b = AC$$, and $$c = AB$$. Then

$$a^2 = BC^2 = (-2+4)^2 + (0-2)^2 = 4 + 4 = 8$$

$$b^2 = AC^2 = (2+4)^2 + (2-2)^2 = 36$$

$$c^2 = AB^2 = (2+2)^2 + (2-0)^2 = 16 + 4 = 20$$

The area of the triangle can be found by the determinant formula: $$\text{Area} = \frac{1}{2}|x_A(y_B - y_C) + x_B(y_C - y_A) + x_C(y_A - y_B)|$$, which becomes $$\frac{1}{2}|2(0-2) + (-2)(2-2) + (-4)(2-0)| = 6$$.

Finally, the circumradius is given by $$R = \frac{abc}{4 \cdot \text{Area}}$$, so

$$R^2 = \frac{a^2 b^2 c^2}{16 \cdot \text{Area}^2} = \frac{8 \times 36 \times 20}{16 \times 36} = \frac{5760}{576} = 10$$

Therefore, $$r^2 = \boxed{10}$$.

Let the side lengths opposite the angles $$A,B,C$$ (with $$A\lt B\lt C$$) be $$a,b,c$$ respectively. They are in an arithmetic progression, so $$a+b=c+b-a\;\Rightarrow\;b-a=c-b.$$

Because the vertices lie on a circle of radius $$R=1$$, the sine rule gives $$\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=2R=2.$$ Hence $$a=2\sin A,\; b=2\sin B,\; c=2\sin C\quad -(1).$$

The largest angle exceeds the smallest by $$\dfrac{\pi}{2}$$, so $$C-A=\frac{\pi}{2}\;\Longrightarrow\;C=A+\frac{\pi}{2}\quad -(2).$$

Using the A.P. condition with $$(1)$$, $$\sin B-\sin A=\sin C-\sin B\;\Longrightarrow\;2\sin B=\sin A+\sin C\quad -(3).$$

Substitute $$C=A+\dfrac{\pi}{2}$$ in $$(3)$$. Because $$\sin\!\left(A+\dfrac{\pi}{2}\right)=\cos A$$ and $$\sin B=\sin\!\left(\dfrac{\pi}{2}-2A\right)=\cos 2A$$, equation $$(3)$$ becomes $$\cos 2A=\frac{\sin A+\cos A}{2}\quad -(4).$$

Write $$\sin A=s,\; \cos A=c$$ so that $$s^2+c^2=1$$. From $$(4)$$, $$c^2-s^2=\dfrac{s+c}{2}$$. Re-arranging: $$2(c-s)(c+s)=s+c\;\Rightarrow\;2(c-s)-1=0\;(\text{since }s+c\neq0).$$ Thus $$c-s=\frac12\quad -(5).$$

With $$c^2+s^2=1$$ and $$(5)$$, $$cs=\frac12\!\Bigl(1-(c-s)^2\Bigr)=\frac12\!\left(1-\frac14\right)=\frac38.$$ Solving the pair $$\begin{cases}c-s=\dfrac12\\[4pt]cs=\dfrac38\end{cases}$$ gives the acute solution $$\sin A=\dfrac{\sqrt7-1}{4},\qquad \cos A=\dfrac{\sqrt7+1}{4}.$$

Consequently $$\sin B=\cos 2A=c^2-s^2=\frac{\sqrt7}{4},\qquad \sin C=\cos A=\frac{\sqrt7+1}{4}.$$

Applying $$(1)$$, the sides are $$a=2\sin A=\frac{\sqrt7-1}{2},\; b=2\sin B=\frac{\sqrt7}{2},\; c=2\sin C=\frac{\sqrt7+1}{2},$$ verifying the arithmetic progression.

Using the formula $$\Delta=\dfrac{abc}{4R}$$ with $$R=1$$, $$\Delta=\frac{1}{4}\left(\frac{\sqrt7-1}{2}\right)\!\left(\frac{\sqrt7}{2}\right)\!\left(\frac{\sqrt7+1}{2}\right) =\frac{3\sqrt7}{16}.$$

Semiperimeter $$s=\frac{a+b+c}{2} =\frac{\tfrac{\sqrt7-1}{2}+\tfrac{\sqrt7}{2}+\tfrac{\sqrt7+1}{2}}{2} =\frac{3\sqrt7}{4}.$$

The inradius is $$r=\frac{\Delta}{s} =\frac{\tfrac{3\sqrt7}{16}}{\tfrac{3\sqrt7}{4}} =\frac14=0.25.$$

Therefore, the required inradius is 0.25.

Let the centre of the big circle $$G$$ be $$O$$ and its radius be $$R$$.
Each small circle $$G_i$$ has centre $$A_i$$ and radius $$r$$.

Because every $$G_i$$ touches $$G$$ externally, the distance $$OA_i$$ is the sum of the radii: $$OA_i = R + r$$.
Because consecutive circles $$G_i$$ and $$G_{i+1}$$ touch externally, the distance between their centres is $$A_iA_{i+1}=2r$$.

The centres $$A_1,A_2,\ldots ,A_n$$ are therefore the vertices of a regular $$n$$-gon inscribed in the circle with centre $$O$$ and radius $$R+r$$.
Hence the central angle between two consecutive vertices is $$\theta = \dfrac{2\pi}{n}$$.

Consider the isosceles triangle $$OA_iA_{i+1}$$. Applying the cosine rule,

$$\begin{aligned} A_iA_{i+1}^2 &= OA_i^2 + OA_{i+1}^2 - 2(OA_i)(OA_{i+1})\cos\theta \\ (2r)^2 &= (R+r)^2 + (R+r)^2 - 2(R+r)^2\cos\theta \\ 4r^2 &= 2(R+r)^2\bigl(1-\cos\theta\bigr). \end{aligned}$$

Rearranging gives

$$\begin{aligned} (R+r)^2 &= \frac{2r^2}{1-\cos\theta},\\ R+r &= \frac{\sqrt{2}\,r}{\sqrt{1-\cos\theta}},\\ R &= r\!\left(\frac{\sqrt{2}}{\sqrt{1-\cos\theta}} - 1\right).\qquad -(1) \end{aligned}$$

Thus the ratio $$\dfrac{R}{r}$$ depends only on $$n$$ through $$\theta=\dfrac{2\pi}{n}$$.

$$\dfrac{R}{r}= \sqrt{2}-1,\qquad\text{so } R = (\sqrt{2}-1)\,r.$$ Option A claims $$(\sqrt{2}-1)r \lt R$$, but we have equality; therefore Option A is false.

$$\dfrac{R}{r}= \frac{\sqrt{2}}{\sqrt{1-0.3090}}-1 \approx \frac{1.4142}{0.8312}-1 \approx 0.698.$$ Hence $$R\approx 0.698\,r,$$ i.e. $$R \lt r.$$ Option B states $$r \lt R$$, which is false.

$$\dfrac{R}{r}= \frac{\sqrt{2}}{\sqrt{1-0.7071}}-1 = \frac{1.4142}{\sqrt{0.2929}}-1 = \frac{1.4142}{0.5412}-1 \approx 1.613.$$ Therefore $$R \approx 1.613\,r \gt (\sqrt{2}-1)\,r\;(\approx 0.414\,r).$$ So the inequality $$(\sqrt{2}-1)r \lt R$$ holds. Option C is true.

$$\dfrac{R}{r}= \frac{\sqrt{2}}{\sqrt{1-0.8660}}-1 = \frac{1.4142}{\sqrt{0.1340}}-1 = \frac{1.4142}{0.3660}-1 \approx 2.866.$$ Compute $$\sqrt{2}(\sqrt{3}+1) = 1.4142\,(1.732 + 1) = 1.4142 \times 2.732 \approx 3.866.$$ Since $$3.866\,r \gt 2.866\,r = R,$$ we have $$\sqrt{2}(\sqrt{3}+1)r \gt R.$$ Option D is true.

Hence the correct statements are:
Option C (If $$n = 8$$, then $$(\sqrt{2}-1)r \lt R$$)
Option D (If $$n = 12$$, then $$\sqrt{2}(\sqrt{3}+1)r \gt R$$).

We need to find the equation of circle $$C$$ that touches the parallel lines $$L_1$$ and $$L_2$$. To determine the constants, note that point $$(-1, 2)$$ lies on $$L_1: 4x - 3y + K_1 = 0$$, so $$4(-1) - 3(2) + K_1 = 0 \implies K_1 = 10$$. Similarly, point $$(3, -6)$$ lies on $$L_2: 4x - 3y + K_2 = 0$$, so $$4(3) - 3(-6) + K_2 = 0 \implies K_2 = -30$$. Hence the equations are $$L_1: 4x - 3y + 10 = 0$$ and $$L_2: 4x - 3y - 30 = 0$$.

Since the circle touches both lines, its centre is equidistant from $$L_1$$ and $$L_2$$ and therefore lies on their midline, given by $$4x - 3y + \frac{10 + (-30)}{2} = 0 \implies 4x - 3y - 10 = 0$$.

The centre also lies on the line through $$(-1, 2)$$ and $$(3, -6)$$, which in parametric form is $$x = -1 + 4t, \quad y = 2 - 8t$$. Substituting into the midline equation $$4x - 3y - 10 = 0$$ gives $$4(-1 + 4t) - 3(2 - 8t) - 10 = 0$$, which simplifies to $$-4 + 16t - 6 + 24t - 10 = 0 \implies 40t - 20 = 0 \implies t = \tfrac{1}{2}$$. Thus the centre is $$\bigl(-1 + 4\cdot\tfrac{1}{2},\;2 - 8\cdot\tfrac{1}{2}\bigr) = (1, -2)\!.$$

The radius is the perpendicular distance from the centre $$(1, -2)$$ to $$L_1$$: $$r = \frac{\bigl|4(1) - 3(-2) + 10\bigr|}{\sqrt{16 + 9}} = \frac{|4 + 6 + 10|}{5} = \frac{20}{5} = 4$$. Verification with $$L_2$$ yields $$\frac{\bigl|4(1) - 3(-2) - 30\bigr|}{5} = 4$$ ✓.

Therefore, the equation of the circle is $$(x - 1)^2 + (y + 2)^2 = 4^2 = 16$$ and the correct answer is Option B: $$(x - 1)^2 + (y + 2)^2 = 16$$.

We have the circle $$x^2 + y^2 - 4x + 3 = 0$$. Rewriting in standard form: $$(x-2)^2 + y^2 = 1$$. So the centre is $$C(2, 0)$$ and radius $$r = 1$$.

The tangents from the origin $$O(0,0)$$ touch the circle at points A and B. The distance from the origin to the centre is $$OC = 2$$.

The length of the tangent from O to the circle is $$OA = OB = \sqrt{OC^2 - r^2} = \sqrt{4 - 1} = \sqrt{3}$$.

Now, in triangle OAC, we have $$OC = 2$$, $$CA = r = 1$$, and $$OA = \sqrt{3}$$. The angle $$\angle AOC$$ can be found using $$\sin(\angle AOC) = \frac{CA}{OC}$$... Actually, since OA is tangent to the circle at A, the angle $$\angle OAC = 90°$$.

So $$\sin(\angle AOC) = \frac{CA}{OC} = \frac{1}{2}$$, giving $$\angle AOC = 30°$$.

By symmetry, $$\angle BOC = 30°$$ as well, so $$\angle AOB = 60°$$.

The area of triangle OAB can be computed as:

Hence, the correct answer is Option 2.

A point P(x, y) moves such that the sum of squares of its distances from (1, 2) and (-2, 1) is 14.

$$(x-1)^2 + (y-2)^2 + (x+2)^2 + (y-1)^2 = 14$$

Expanding each term:

$$x^2 - 2x + 1 + y^2 - 4y + 4 + x^2 + 4x + 4 + y^2 - 2y + 1 = 14$$

$$2x^2 + 2y^2 + 2x - 6y + 10 = 14$$

$$x^2 + y^2 + x - 3y - 2 = 0$$

Rewriting in standard form by completing the square:

$$\left(x + \frac{1}{2}\right)^2 - \frac{1}{4} + \left(y - \frac{3}{2}\right)^2 - \frac{9}{4} - 2 = 0$$

$$\left(x + \frac{1}{2}\right)^2 + \left(y - \frac{3}{2}\right)^2 = \frac{1}{4} + \frac{9}{4} + 2 = \frac{10}{4} + 2 = \frac{9}{2}$$

This is a circle with centre $$\left(-\frac{1}{2}, \frac{3}{2}\right)$$ and radius $$r = \sqrt{\frac{9}{2}} = \frac{3}{\sqrt{2}}$$.

Setting $$y = 0$$: $$x^2 + x - 2 = 0 \implies (x+2)(x-1) = 0$$

Points: $$A(-2, 0)$$ and $$B(1, 0)$$. So $$|AB| = 3$$.

Setting $$x = 0$$: $$y^2 - 3y - 2 = 0$$

$$y = \frac{3 \pm \sqrt{9 + 8}}{2} = \frac{3 \pm \sqrt{17}}{2}$$

Points: $$C\left(0, \frac{3+\sqrt{17}}{2}\right)$$ and $$D\left(0, \frac{3-\sqrt{17}}{2}\right)$$. So $$|CD| = \sqrt{17}$$.

The quadrilateral ACBD has its diagonals along the x-axis (AB) and y-axis (CD), which are perpendicular to each other. For a quadrilateral with perpendicular diagonals:

$$\text{Area} = \frac{1}{2} \times |AB| \times |CD| = \frac{1}{2} \times 3 \times \sqrt{17} = \frac{3\sqrt{17}}{2}$$

Therefore, the correct answer is Option B: $$\dfrac{3\sqrt{17}}{2}$$.

The circle is $$x^2 + y^2 - 2x - 4y = 0$$, which has center $$(1, 2)$$ and radius $$r = \sqrt{1 + 4} = \sqrt{5}$$.

First, we verify that the points $$O(0,0)$$ and $$P(1+\sqrt{5},2)$$ lie on this circle. For $$O(0,0)$$, substitution gives $$0 + 0 - 0 - 0 = 0$$ ✓. For $$P(1+\sqrt{5},2)$$, we compute $$(1+\sqrt{5})^2 + 4 - 2(1+\sqrt{5}) - 8 = 1 + 2\sqrt{5} + 5 + 4 - 2 - 2\sqrt{5} - 8 = 0$$ ✓.

The tangent to the circle $$x^2 + y^2 - 2x - 4y = 0$$ at a general point $$(x_1,y_1)$$ is given by $$xx_1 + yy_1 - (x + x_1) - 2(y + y_1) = 0$$. At $$O(0,0)$$ this becomes $$0 + 0 - (x + 0) - 2(y + 0) = 0$$, which simplifies to $$x + 2y = 0\quad\cdots(1)$$.

At $$P(1+\sqrt{5},2)$$ the same formula yields $$x(1+\sqrt{5}) + 2y - (x + 1 + \sqrt{5}) - 2(y + 2) = 0$$ $$x + x\sqrt{5} + 2y - x - 1 - \sqrt{5} - 2y - 4 = 0$$ $$x\sqrt{5} - 5 - \sqrt{5} = 0$$ $$\sqrt{5}(x - 1 - \sqrt{5}) = 0$$ so $$x = 1 + \sqrt{5}\quad\cdots(2)$$.

Solving equations (1) and (2), from (2) we have $$x = 1 + \sqrt{5}$$. Substituting into (1) gives $$1 + \sqrt{5} + 2y = 0\quad\Rightarrow\quad y = -\frac{1 + \sqrt{5}}{2}$$ so the point of intersection is $$Q = \left(1 + \sqrt{5},\; -\frac{1 + \sqrt{5}}{2}\right)\,.$$

Finally, the area of triangle $$OPQ$$ is given by Area $$= \frac{1}{2}\bigl|x_O(y_P - y_Q) + x_P(y_Q - y_O) + x_Q(y_O - y_P)\bigr|$$ $$= \frac{1}{2}\left|0 + (1+\sqrt{5})\Bigl(-\frac{1+\sqrt{5}}{2} - 0\Bigr) + (1+\sqrt{5})(0 - 2)\right|$$ $$= \frac{1}{2}\left|(1+\sqrt{5})\Bigl(-\frac{1+\sqrt{5}}{2} - 2\Bigr)\right|$$ $$= \frac{1}{2}\left|(1+\sqrt{5}) \cdot \Bigl(-\frac{5+\sqrt{5}}{2}\Bigr)\right| = \frac{(1+\sqrt{5})(5+\sqrt{5})}{4}\,.$$ Expanding the numerator, $$(1+\sqrt{5})(5+\sqrt{5}) = 5 + \sqrt{5} + 5\sqrt{5} + 5 = 10 + 6\sqrt{5}\,,$$ so Area $$= \frac{10 + 6\sqrt{5}}{4} = \frac{5 + 3\sqrt{5}}{2}\,.$$ Therefore, the correct answer is Option C: $$\frac{5 + 3\sqrt{5}}{2}\,.$$

We are given that the abscissae of points $$P$$ and $$Q$$ are roots of $$x^2 - 4x - 6 = 0$$ and the ordinates are roots of $$y^2 + 2y - 7 = 0$$. $$PQ$$ is a diameter of the circle $$x^2 + y^2 + 2ax + 2by + c = 0$$.

Since $$PQ$$ is a diameter, the center of the circle is the midpoint of $$PQ$$.

Let $$P = (x_1, y_1)$$ and $$Q = (x_2, y_2)$$.

From $$x^2 - 4x - 6 = 0$$: $$x_1 + x_2 = 4$$ (sum of roots)

From $$y^2 + 2y - 7 = 0$$: $$y_1 + y_2 = -2$$ (sum of roots)

Center = $$\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right) = (2, -1)$$

Comparing with the general form $$x^2 + y^2 + 2ax + 2by + c = 0$$, the center is $$(-a, -b)$$.

$$-a = 2 \Rightarrow a = -2$$

$$-b = -1 \Rightarrow b = 1$$

$$|PQ|^2 = (x_1-x_2)^2 + (y_1-y_2)^2$$

$$(x_1-x_2)^2 = (x_1+x_2)^2 - 4x_1x_2 = 16 - 4(-6) = 16 + 24 = 40$$

$$(y_1-y_2)^2 = (y_1+y_2)^2 - 4y_1y_2 = 4 - 4(-7) = 4 + 28 = 32$$

$$|PQ|^2 = 40 + 32 = 72$$

Radius$$^2 = \frac{|PQ|^2}{4} = \frac{72}{4} = 18$$

For the circle $$x^2 + y^2 + 2ax + 2by + c = 0$$:

$$r^2 = a^2 + b^2 - c$$

$$18 = (-2)^2 + 1^2 - c = 4 + 1 - c = 5 - c$$

$$c = 5 - 18 = -13$$

$$a + b - c = -2 + 1 - (-13) = -2 + 1 + 13 = 12$$

Therefore, the correct answer is Option A: $$12$$.

We are given an equilateral triangle ABC with vertices $$A(\sin t,-\cos t)$$, $$B(\cos t,\sin t)$$, and $$C(a,b)$$, and its orthocentre lies on a circle with centre $$(1,\tfrac13)$$.

Since the triangle is equilateral, its orthocentre coincides with its centroid, which is

$$G=\Bigl(\frac{\sin t+\cos t+a}{3},\;\frac{-\cos t+\sin t+b}{3}\Bigr).$$

Letting $$G=(X,Y)$$, we have

$$X=\frac{\sin t+\cos t+a}{3},\qquad Y=\frac{\sin t-\cos t+b}{3},$$

so that

$$3X-a=\sin t+\cos t,\qquad 3Y-b=\sin t-\cos t.$$

Squaring and adding these two equations yields

$$ (3X-a)^2+(3Y-b)^2=(\sin t+\cos t)^2+(\sin t-\cos t)^2, $$

which simplifies to

$$\sin^2t+2\sin t\cos t+\cos^2t+\sin^2t-2\sin t\cos t+\cos^2t=2. $$

This shows that the centroid traces the circle

$$ (3X-a)^2+(3Y-b)^2=2, $$

or equivalently

$$\Bigl(X-\frac{a}{3}\Bigr)^2+\Bigl(Y-\frac{b}{3}\Bigr)^2=\frac{2}{9}.$$

From the above equation, the centre of this circle is $$\bigl(\tfrac{a}{3},\tfrac{b}{3}\bigr)$$, which must coincide with the given centre $$(1,\tfrac13)$$. Therefore,

$$\frac{a}{3}=1\;\Rightarrow\;a=3,\qquad \frac{b}{3}=\frac{1}{3}\;\Rightarrow\;b=1.$$

Substituting these values gives

$$a^2-b^2=9-1=8.$$

The correct answer is Option B: $$8$$.

Circle $$C_1$$ passes through the origin $$O(0,0)$$ and has diameter 4 on the positive $$x$$-axis. Its centre is $$(2, 0)$$, radius $$= 2$$, so its equation is $$(x-2)^2 + y^2 = 4$$.

The line $$y = 2x$$ intersects $$C_1$$; substituting gives $$(x-2)^2 + 4x^2 = 4 \implies 5x^2 - 4x = 0 \implies x(5x - 4) = 0,$$ so $$x = 0$$ (point $$O$$) or $$x = \dfrac{4}{5}$$. Hence $$A = \left(\dfrac{4}{5}, \dfrac{8}{5}\right)$$.

The centre of $$C_2$$ is the midpoint of $$OA$$, namely $$\left(\dfrac{2}{5}, \dfrac{4}{5}\right)$$, and its radius is half of $$|OA|$$: $$\dfrac{|OA|}{2} = \dfrac{1}{2}\sqrt{\dfrac{16}{25}+\dfrac{64}{25}} = \dfrac{1}{2} \cdot \dfrac{\sqrt{80}}{5} = \dfrac{2\sqrt{5}}{5}$$.

For a circle with centre $$(h,k)$$, the tangent at $$(x_1, y_1)$$ satisfies $$(x_1 - h)(x - h) + (y_1 - k)(y - k) = r^2.$$ Substituting $$h=\frac{2}{5}$$, $$k=\frac{4}{5}$$, $$(x_1,y_1)=\left(\frac{4}{5},\frac{8}{5}\right)$$ and $$r=\frac{2\sqrt{5}}{5}$$ gives $$\left(\frac{4}{5}-\frac{2}{5}\right)\left(x-\frac{2}{5}\right) + \left(\frac{8}{5}-\frac{4}{5}\right)\left(y-\frac{4}{5}\right) = \frac{4}{5},$$ which simplifies to $$\frac{2}{5}\left(x-\frac{2}{5}\right) + \frac{4}{5}\left(y-\frac{4}{5}\right) = \frac{4}{5}.$$ Multiplying by 5 yields $$2\left(x-\dfrac{2}{5}\right) + 4\left(y-\dfrac{4}{5}\right) = 4,$$ so $$2x - \frac{4}{5} + 4y - \frac{16}{5} = 4 \implies 2x + 4y = 8 \implies x + 2y = 4.$$

Setting $$y=0$$ gives $$x=4$$ so $$P = (4,0)$$, and setting $$x=0$$ gives $$2y=4$$ so $$y=2$$ and $$Q=(0,2)$$.

Then $$QA = \sqrt{\left(\frac{4}{5}\right)^2 + \left(\frac{8}{5}-2\right)^2} = \sqrt{\frac{16}{25}+\frac{4}{25}} = \frac{\sqrt{20}}{5} = \frac{2\sqrt{5}}{5},$$ $$AP = \sqrt{\left(4-\frac{4}{5}\right)^2 + \left(\frac{8}{5}\right)^2} = \sqrt{\frac{256}{25}+\frac{64}{25}} = \frac{\sqrt{320}}{5} = \frac{8\sqrt{5}}{5},$$ so $$QA : AP = \frac{2\sqrt{5}}{5} : \frac{8\sqrt{5}}{5} = 2 : 8 = \boxed{1 : 4}.$$

The answer is Option A.

We are given a circle $$x^2 - \sqrt{2}(x + y) + y^2 = 0$$ with an inscribed triangle $$ABC$$ where $$\angle BAC = \frac{\pi}{2}$$ and $$AB = \sqrt{2}$$.

Rewriting the circle equation in standard form, $$x^2 - \sqrt{2}x + y^2 - \sqrt{2}y = 0$$ so that $$(x - \frac{\sqrt{2}}{2})^2 + (y - \frac{\sqrt{2}}{2})^2 = \frac{1}{2} + \frac{1}{2} = 1$$ and hence the center is $$\left(\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}\right)$$ with radius $$R = 1$$.

Since $$\angle BAC = \frac{\pi}{2}$$, the chord $$BC$$ subtends a right angle at a point on the circle and by the inscribed angle theorem must be a diameter, so $$BC = 2R = 2$$.

Triangle $$ABC$$ is right-angled at $$A$$ with $$AB = \sqrt{2}$$ and hypotenuse $$BC = 2$$, and by the Pythagorean theorem $$AC = \sqrt{BC^2 - AB^2} = \sqrt{4 - 2} = \sqrt{2}$$.

Therefore, the area is $$\frac{1}{2} \times AB \times AC = \frac{1}{2} \times \sqrt{2} \times \sqrt{2} = \frac{1}{2} \times 2 = 1$$.

Therefore, the answer is Option A: $$\textbf{1}$$.

We have the circle $$x^2 + y^2 - x + 2y = \frac{11}{4}$$ and need to find the area of triangle PQR.

First, rewrite the equation by completing the squares: $$\left(x - \frac{1}{2}\right)^2 + (y + 1)^2 = \frac{11}{4} + \frac{1}{4} + 1 = 4$$, which shows that the centre is $$C = \left(\frac{1}{2}, -1\right)$$ and the radius is $$r = 2$$.

Let P be any point on the circle. A line passing through the centre C makes an angle of $$\frac{\pi}{4}$$ with the line CP and intersects the circle again at Q and R. Since this line goes through the centre, QR is a diameter of the circle, so $$|QR| = 2r = 4$$.

To determine the height of triangle PQR from P to QR, observe that the angle between CP and QR is $$\frac{\pi}{4}$$ and that $$|CP| = r = 2$$ because P lies on the circle. The perpendicular distance from P to the line QR is therefore $$h = |CP| \sin\frac{\pi}{4} = 2 \cdot \frac{1}{\sqrt{2}} = \sqrt{2}$$.

Using the base QR and the corresponding height h, the area of triangle PQR is calculated as $$\text{Area} = \frac{1}{2} \times |QR| \times h = \frac{1}{2} \times 4 \times \sqrt{2} = 2\sqrt{2}$$.

Therefore, the correct answer is Option B: $$2\sqrt{2}$$.

The circle $$x^2 + y^2 + Ax + By + C = 0$$ passes through $$(0, 6)$$ and touches the parabola $$y = x^2$$ at $$(2, 4)$$.

Since the circle passes through $$(0, 6)$$:

$$ 0 + 36 + 0 + 6B + C = 0 $$

$$ 6B + C = -36 \quad \cdots (1) $$

Since the circle passes through $$(2, 4)$$:

$$ 4 + 16 + 2A + 4B + C = 0 $$

$$ 2A + 4B + C = -20 \quad \cdots (2) $$

The circle touches the parabola $$y = x^2$$ at $$(2, 4)$$, so they have the same tangent at that point.

The slope of the parabola at $$(2, 4)$$: $$\frac{dy}{dx} = 2x = 4$$.

For the circle, differentiating $$x^2 + y^2 + Ax + By + C = 0$$ implicitly:

$$ 2x + 2y\frac{dy}{dx} + A + B\frac{dy}{dx} = 0 $$

At $$(2, 4)$$ with $$\frac{dy}{dx} = 4$$:

$$ 4 + 8(4) + A + 4B = 0 $$

$$ 4 + 32 + A + 4B = 0 $$

$$ A + 4B = -36 \quad \cdots (3) $$

Solve the system of equations.

From (3): $$A = -36 - 4B$$

Substitute into (2):

$$ 2(-36 - 4B) + 4B + C = -20 $$

$$ -72 - 8B + 4B + C = -20 $$

$$ -4B + C = 52 \quad \cdots (4) $$

From (1): $$C = -36 - 6B$$. Substitute into (4):

$$ -4B + (-36 - 6B) = 52 $$

$$ -10B - 36 = 52 $$

$$ -10B = 88 $$

$$ B = -\frac{88}{10} = -\frac{44}{5} $$

$$ C = -36 - 6\left(-\frac{44}{5}\right) = -36 + \frac{264}{5} = \frac{-180 + 264}{5} = \frac{84}{5} $$

$$ A = -36 - 4\left(-\frac{44}{5}\right) = -36 + \frac{176}{5} = \frac{-180 + 176}{5} = -\frac{4}{5} $$

Compute $$A + C$$:

$$ A + C = -\frac{4}{5} + \frac{84}{5} = \frac{80}{5} = 16 $$

The answer is Option A: 16.

The circle is $$C: 4x^2 + 4y^2 - 12x + 8y + k = 0$$. We need to find $$k$$ such that the circle lies inside the fourth quadrant and the point $$(1, -\frac{1}{3})$$ lies on or inside the circle.

Rewrite the circle in standard form

Dividing by 4: $$x^2 + y^2 - 3x + 2y + \frac{k}{4} = 0$$

Completing the square:

$$\left(x - \frac{3}{2}\right)^2 + (y + 1)^2 = \frac{9}{4} + 1 - \frac{k}{4} = \frac{13 - k}{4}$$

Centre: $$\left(\frac{3}{2}, -1\right)$$, Radius: $$r = \frac{\sqrt{13 - k}}{2}$$

Condition for the circle to exist

$$r^2 > 0 \Rightarrow 13 - k > 0 \Rightarrow k < 13$$

Circle lies in the fourth quadrant

The centre is $$\left(\frac{3}{2}, -1\right)$$ which is in the fourth quadrant ($$x > 0, y < 0$$). For the entire circle to lie in the fourth quadrant:

The circle must not cross the x-axis ($$y = 0$$): distance from centre to x-axis $$>$$ radius

$$1 > \frac{\sqrt{13-k}}{2} \Rightarrow 4 > 13 - k \Rightarrow k > 9$$

The circle must not cross the y-axis ($$x = 0$$): distance from centre to y-axis $$>$$ radius

$$\frac{3}{2} > \frac{\sqrt{13-k}}{2} \Rightarrow 9 > 13 - k \Rightarrow k > 4$$

This is already satisfied when $$k > 9$$.

Point $$(1, -\frac{1}{3})$$ lies on or inside the circle

Substituting into the circle equation (the point must satisfy $$\leq 0$$):

$$4(1)^2 + 4\left(\frac{1}{9}\right) - 12(1) + 8\left(-\frac{1}{3}\right) + k \leq 0$$

$$4 + \frac{4}{9} - 12 - \frac{8}{3} + k \leq 0$$

$$\frac{36 + 4 - 108 - 24}{9} + k \leq 0$$

$$\frac{-92}{9} + k \leq 0$$

$$k \leq \frac{92}{9}$$

Combine the conditions

$$k > 9$$ and $$k \leq \frac{92}{9}$$

So $$k \in \left(9, \frac{92}{9}\right]$$

The correct answer is Option D: $$\left(9, \frac{92}{9}\right]$$.

Set up the conditions.

Let the center of the circle be $$(h, k)$$ with radius $$r$$.

Since the circle touches the $$y$$-axis, the distance from the center to the $$y$$-axis equals the radius:

$$r = |h|$$

Since the circle also touches the line $$x + y = 0$$, the distance from the center to this line equals the radius:

$$\frac{|h + k|}{\sqrt{2}} = r = |h|$$

Solve for the relationship.

Squaring both sides:

$$\frac{(h + k)^2}{2} = h^2$$

$$(h + k)^2 = 2h^2$$

$$h^2 + 2hk + k^2 = 2h^2$$

$$k^2 + 2hk - h^2 = 0$$

$$h^2 - k^2 = 2hk$$

Write the locus by replacing $$h$$ with $$x$$ and $$k$$ with $$y$$:

$$x^2 - y^2 = 2xy$$

Therefore, the locus of the center is $$x^2 - y^2 = 2xy$$, which is Option D.

We are given $$\frac{dy}{dx} = \frac{ax - by + a}{bx + cy + a}$$ represents a circle passing through $$(2, 5)$$.

For the ODE to represent a circle, we recall that the general equation of a circle can be written as $$F(x, y) = 0$$. Taking the total differential gives $$F_x + F_y \frac{dy}{dx} = 0$$, so $$\frac{dy}{dx} = -\frac{F_x}{F_y}$$. In the case of a circle with equation $$x^2 + y^2 + 2gx + 2fy + d = 0$$, we have $$F_x = 2x + 2g$$ and $$F_y = 2y + 2f$$, which yields $$\frac{dy}{dx} = -\frac{2x + 2g}{2y + 2f} = -\frac{x + g}{y + f}$$.

Next, comparing the given form $$\frac{dy}{dx} = \frac{ax - by + a}{bx + cy + a}$$ with the expression $$-\frac{x + g}{y + f}$$, we see that there must be no $$y$$ term in the numerator and no $$x$$ term in the denominator, so $$b = 0$$. Thus the numerator becomes $$ax + a = a(x + 1)$$ and the denominator becomes $$cy + a$$. Equating these, we write $$\frac{a(x+1)}{cy + a} = -\frac{x + g}{y + f}$$, which implies for some $$\lambda$$, $$a(x+1) = -\lambda(x + g)\quad\text{and}\quad cy + a = \lambda(y + f)\,.$$ From the first equation it follows that $$a = -\lambda$$ and $$a = -\lambda g$$, hence $$g = 1$$. From the second equation we obtain $$c = \lambda$$ and $$a = \lambda f$$, giving $$c = -a$$ and $$f = -1$$.

Substituting these values into the circle equation yields $$x^2 + y^2 + 2(1)x + 2(-1)y + d = 0\,, $$ or equivalently $$x^2 + y^2 + 2x - 2y + d = 0\,. $$ Since the circle passes through $$(2, 5)$$, we substitute to find $$4 + 25 + 4 - 10 + d = 0 \quad\Longrightarrow\quad d = -23\,. $$ Hence the required circle is $$x^2 + y^2 + 2x - 2y - 23 = 0\,. $$

The center of this circle is $$(-1, 1)$$ and its radius is $$\sqrt{1 + 1 + 23} = \sqrt{25} = 5\,. $$ For the point $$(11,6)$$, the distance to the center $$(-1,1)$$ is $$\sqrt{(11-(-1))^2 + (6-1)^2} = \sqrt{144 + 25} = \sqrt{169} = 13\,, $$ so the shortest distance from $$(11,6)$$ to the circle is $$13 - 5 = 8\,. $$

The answer is Option B: $$8$$.

Circle 1: $$x^2 + y^2 + 6x + 8y + 16 = 0$$ has center $$C_1 = (-3,-4)$$ and radius $$r_1 = \sqrt{9 + 16 - 16} = 3$$.

Circle 2: $$x^2 + y^2 + 2(3-\sqrt{3})x + 2(4-\sqrt{6})y = k + 6\sqrt{3} + 8\sqrt{6}$$ has center $$C_2 = \bigl(-(3-\sqrt{3}),\,-(4-\sqrt{6})\bigr) = (-3+\sqrt{3}, -4+\sqrt{6})$$.

For the radius $$r_2$$ of the second circle we have

The distance between the centers is

For internal tangency we set

Since $$k > 0$$ we have $$\sqrt{34+k} = 6\implies k = 2$$, so $$r_2 = 6$$.

The point of tangency $$P(\alpha,\beta)$$ lies on the line $$C_1C_2$$ at distance $$r_1 = 3$$ from $$C_1$$ in the direction opposite to $$C_2$$, hence

Finally,

The answer is $$\boxed{25}$$.

A circle $$C$$ of radius 5 lies below the $$x$$-axis. The line $$L_1: 4x + 3y + 2 = 0$$ passes through the centre $$P$$, while $$L_2: 3x - 4y - 11 = 0$$ is tangent to $$C$$ at the point $$Q$$, which is their intersection.

To find $$Q$$, solve $$4x + 3y = -2$$ and $$3x - 4y = 11$$ simultaneously. Multiplying the first equation by 4 and the second by 3 gives $$16x + 12y = -8$$ and $$9x - 12y = 33$$, and adding these yields $$25x = 25$$ so $$x = 1$$ and hence $$y = -2$$. Thus $$Q = (1, -2)$$.

Since $$L_2$$ is tangent to the circle at $$Q$$, the line $$PQ$$ is perpendicular to $$L_2$$. The normal vector to $$L_2: 3x - 4y - 11 = 0$$ is $$(3, -4)$$ with unit vector $$\frac{1}{5}(3, -4)$$. Therefore

$$P = Q \pm 5 \cdot \frac{(3, -4)}{5} = (1, -2) \pm (3, -4),$$

giving $$P = (4, -6)\text{ or }P = (-2, 2).$$

Because the circle lies below the $$x$$-axis, the centre must have $$y < 0$$, so $$P = (4, -6)$$. Verifying on $$L_1$$: $$4(4) + 3(-6) + 2 = 16 - 18 + 2 = 0$$.

Finally, the distance from $$P$$ to the line $$5x - 12y + 51 = 0$$ is

$$d = \frac{\lvert 5(4) - 12(-6) + 51 \rvert}{\sqrt{25 + 144}} = \frac{\lvert 20 + 72 + 51 \rvert}{13} = \frac{143}{13} = 11.$$

The correct answer is 11.

A rectangle $$R$$ has one side with endpoints $$(1, 2)$$ and $$(3, 6)$$, is inscribed in a circle, and the circle has diameter $$2x - y + 4 = 0$$. Find the area of $$R$$.

The center of the circle lies on the diameter $$2x - y + 4 = 0$$ and, since for a rectangle inscribed in a circle the center is the midpoint of the diagonals and is equidistant from all four vertices, we proceed by finding key points.

The midpoint of $$(1, 2)$$ and $$(3, 6)$$ is $$M = (2, 4)$$.

Because the perpendicular bisector of any chord passes through the center, we compute the slope of the side from $$(1, 2)$$ to $$(3, 6)$$ as $$m = \frac{6-2}{3-1} = 2$$. Hence the perpendicular bisector has slope $$-\frac{1}{2}$$ and passes through $$(2, 4)$$. Its equation is $$y - 4 = -\frac{1}{2}(x - 2)$$, which simplifies to $$2y - 8 = -x + 2$$ and therefore $$x + 2y = 10$$.

Substituting $$2x - y + 4 = 0 \Rightarrow y = 2x + 4$$ into $$x + 2y = 10$$ gives $$x + 2(2x + 4) = 10$$, so $$x + 4x + 8 = 10$$ and thus $$5x = 2$$. Therefore, $$x = \frac{2}{5}, \quad y = 2 \cdot \frac{2}{5} + 4 = \frac{24}{5}$$ and the center is $$\left(\frac{2}{5}, \frac{24}{5}\right)$$.

It follows that the radius satisfies $$r^2 = \left(1 - \frac{2}{5}\right)^2 + \left(2 - \frac{24}{5}\right)^2 = \left(\frac{3}{5}\right)^2 + \left(-\frac{14}{5}\right)^2 = \frac{9}{25} + \frac{196}{25} = \frac{205}{25} = \frac{41}{5}$$.

The length of the given side is $$|AB| = \sqrt{(3-1)^2 + (6-2)^2} = \sqrt{4 + 16} = \sqrt{20} = 2\sqrt{5}$$.

Since the diagonal of the rectangle equals the diameter, we have diameter = $$2r$$, so diagonal$$^2 = 4r^2 = \frac{4 \times 41}{5} = \frac{164}{5}$$. For a rectangle: diagonal$$^2$$ = side$$_1^2$$ + side$$_2^2$$, hence $$\frac{164}{5} = 20 + \text{side}_2^2$$, which gives $$\text{side}_2^2 = \frac{164}{5} - 20 = \frac{164 - 100}{5} = \frac{64}{5}$$ and $$\text{side}_2 = \frac{8}{\sqrt{5}}$$.

Therefore, the area of the rectangle is $$2\sqrt{5} \times \frac{8}{\sqrt{5}} = 16$$.

The answer is $$\boxed{16}$$.

The circle $$C: (x-h)^2 + (y-k)^2 = r^2$$ with $$k > 0$$ touches the x-axis at $$(1, 0)$$.

Since the circle touches the x-axis at $$(1, 0)$$, the center must be directly above this point, so $$h = 1$$ and $$k = r$$ (since the distance from center to x-axis equals the radius).

The circle is: $$(x-1)^2 + (y-r)^2 = r^2$$

The line $$x + y = 0$$ intersects the circle such that chord $$PQ$$ has length 2.

The distance from the center $$(1, r)$$ to the line $$x + y = 0$$ is:

$$d = \frac{|1 + r|}{\sqrt{2}} = \frac{1 + r}{\sqrt{2}}$$ (since $$r > 0$$)

Using the chord length formula: $$\left(\frac{PQ}{2}\right)^2 = r^2 - d^2$$

$$1 = r^2 - \frac{(1+r)^2}{2}$$

$$2 = 2r^2 - (1 + r)^2$$

$$2 = 2r^2 - 1 - 2r - r^2$$

$$2 = r^2 - 2r - 1$$

$$r^2 - 2r - 3 = 0$$

$$(r - 3)(r + 1) = 0$$

Since $$r > 0$$, we get $$r = 3$$.

Therefore: $$h + k + r = 1 + 3 + 3 = 7$$

The correct answer is $$7$$.

We have the circle $$(x-2)^2 + (y+1)^2 = \frac{169}{4}$$ with center $$C = (2, -1)$$ and radius $$r = \frac{13}{2}$$. A chord $$AB$$ has length 12, and tangents drawn to the circle at points $$A$$ and $$B$$ intersect at point $$P$$. We need to find five times the distance from $$P$$ to chord $$AB$$.

Let $$M$$ be the midpoint of chord $$AB$$. The perpendicular from the center of a circle to a chord bisects the chord, so $$CM \perp AB$$ and $$AM = MB = 6$$. Using the Pythagorean theorem in right triangle $$CMA$$: $$CM = \sqrt{CA^2 - AM^2} = \sqrt{r^2 - 36} = \sqrt{\frac{169}{4} - 36} = \sqrt{\frac{169 - 144}{4}} = \sqrt{\frac{25}{4}} = \frac{5}{2}$$.

By the symmetry of the configuration ($$A$$ and $$B$$ are symmetric about line $$CM$$), the intersection point $$P$$ of the tangents at $$A$$ and $$B$$ lies on line $$CM$$, on the opposite side of the chord from the center. The distance from $$P$$ to the chord is $$PM$$, and $$CP = CM + PM = \frac{5}{2} + PM$$.

Since $$PA$$ is tangent to the circle at $$A$$, the radius $$CA$$ is perpendicular to the tangent $$PA$$, so $$\angle CAP = 90°$$. In right triangle $$CAP$$, the Pythagorean theorem gives $$CP^2 = CA^2 + PA^2 = \frac{169}{4} + PA^2$$.

Also, since $$PM \perp AB$$ and $$A$$ lies on chord $$AB$$, triangle $$PMA$$ is right-angled at $$M$$. So $$PA^2 = PM^2 + MA^2 = PM^2 + 36$$.

Substituting into the equation from triangle $$CAP$$: $$\left(\frac{5}{2} + PM\right)^2 = \frac{169}{4} + PM^2 + 36$$.

Expanding the left side: $$\frac{25}{4} + 5 \cdot PM + PM^2 = \frac{169}{4} + PM^2 + 36$$.

The $$PM^2$$ terms cancel, leaving $$\frac{25}{4} + 5 \cdot PM = \frac{169}{4} + 36$$. Computing the right side: $$\frac{169}{4} + \frac{144}{4} = \frac{313}{4}$$.

So $$5 \cdot PM = \frac{313}{4} - \frac{25}{4} = \frac{288}{4} = 72$$, giving $$PM = \frac{72}{5}$$.

Therefore, five times the distance from $$P$$ to chord $$AB$$ is $$5 \times PM = 5 \times \frac{72}{5} = 72$$.

Hence, the correct answer is $$\boxed{72}$$.

The center of the circle $$(h, k)$$ lies on both normals:

$$k + 2h = \sqrt{11} + 7\sqrt{7} \quad \cdots (1)$$

$$2k + h = 2\sqrt{11} + 6\sqrt{7} \quad \cdots (2)$$

From (1): $$k = \sqrt{11} + 7\sqrt{7} - 2h$$. Substituting into (2):

$$2(\sqrt{11} + 7\sqrt{7} - 2h) + h = 2\sqrt{11} + 6\sqrt{7}$$

$$2\sqrt{11} + 14\sqrt{7} - 4h + h = 2\sqrt{11} + 6\sqrt{7}$$

$$-3h = -8\sqrt{7}$$

$$h = \frac{8\sqrt{7}}{3}$$

$$k = \sqrt{11} + 7\sqrt{7} - \frac{16\sqrt{7}}{3} = \sqrt{11} + \frac{21\sqrt{7} - 16\sqrt{7}}{3} = \sqrt{11} + \frac{5\sqrt{7}}{3}$$

The tangent line is $$\sqrt{11}y - 3x = \frac{5\sqrt{77}}{3} + 11$$, or $$3x - \sqrt{11}y + \frac{5\sqrt{77}}{3} + 11 = 0$$.

Multiply through by 3: $$9x - 3\sqrt{11}y + 5\sqrt{77} + 33 = 0$$.

The distance from center $$(h, k)$$ to this line equals $$r$$:

$$r = \frac{|9h - 3\sqrt{11}k + 5\sqrt{77} + 33|}{\sqrt{81 + 99}} = \frac{|9h - 3\sqrt{11}k + 5\sqrt{77} + 33|}{\sqrt{180}} = \frac{|9h - 3\sqrt{11}k + 5\sqrt{77} + 33|}{6\sqrt{5}}$$

Computing $$9h - 3\sqrt{11}k$$:

$$9h = 9 \cdot \frac{8\sqrt{7}}{3} = 24\sqrt{7}$$

$$3\sqrt{11}k = 3\sqrt{11}\left(\sqrt{11} + \frac{5\sqrt{7}}{3}\right) = 3 \cdot 11 + 5\sqrt{77} = 33 + 5\sqrt{77}$$

$$9h - 3\sqrt{11}k + 5\sqrt{77} + 33 = 24\sqrt{7} - 33 - 5\sqrt{77} + 5\sqrt{77} + 33 = 24\sqrt{7}$$

$$r = \frac{24\sqrt{7}}{6\sqrt{5}} = \frac{4\sqrt{7}}{\sqrt{5}}$$

$$r^2 = \frac{112}{5}$$

Now computing $$(5h - 8k)^2 + 5r^2$$:

$$5h = \frac{40\sqrt{7}}{3}, \quad 8k = 8\sqrt{11} + \frac{40\sqrt{7}}{3}$$

$$5h - 8k = \frac{40\sqrt{7}}{3} - 8\sqrt{11} - \frac{40\sqrt{7}}{3} = -8\sqrt{11}$$

$$(5h - 8k)^2 = 64 \times 11 = 704$$

$$5r^2 = 5 \times \frac{112}{5} = 112$$

$$(5h - 8k)^2 + 5r^2 = 704 + 112 = 816$$

Hence the answer is $$\boxed{816}$$.

The circle $$c_1: x^2 + y^2 - 2x - 6y + \alpha = 0$$ has centre $$(1, 3)$$ and radius $$r_1 = \sqrt{1 + 9 - \alpha} = \sqrt{10 - \alpha}$$.

We reflect the centre $$(1, 3)$$ in the line $$y = x + 1$$, i.e., $$x - y + 1 = 0$$. The reflection of a point $$(h, k)$$ in the line $$ax + by + c = 0$$ is given by $$\left(h - \frac{2a(ah+bk+c)}{a^2+b^2},\; k - \frac{2b(ah+bk+c)}{a^2+b^2}\right)$$. With $$a = 1, b = -1, c = 1$$:

$$ah + bk + c = 1 - 3 + 1 = -1$$, and $$a^2 + b^2 = 2$$.

The reflected centre is $$\left(1 - \frac{2(1)(-1)}{2},\; 3 - \frac{2(-1)(-1)}{2}\right) = (1 + 1,\; 3 - 1) = (2, 2)$$.

The reflected circle $$c_2$$ has the same radius as $$c_1$$ and centre $$(2, 2)$$. Its equation is $$(x-2)^2 + (y-2)^2 = r_1^2 = 10 - \alpha$$, which expands to $$x^2 + y^2 - 4x - 4y + 8 - (10-\alpha) = 0$$, i.e., $$x^2 + y^2 - 4x - 4y + \alpha - 2 = 0$$.

We are told $$c_2: 5x^2 + 5y^2 + 10gx + 10fy + 38 = 0$$, which in standard form (dividing by 5) is $$x^2 + y^2 + 2gx + 2fy + \frac{38}{5} = 0$$.

Comparing with $$x^2 + y^2 - 4x - 4y + (\alpha - 2) = 0$$: $$2g = -4 \Rightarrow g = -2$$, $$2f = -4 \Rightarrow f = -2$$, and $$\frac{38}{5} = \alpha - 2 \Rightarrow \alpha = 2 + \frac{38}{5} = \frac{48}{5}$$.

The radius of $$c_2$$ is $$r = \sqrt{g^2 + f^2 - 38/5} = \sqrt{4 + 4 - 38/5} = \sqrt{\frac{40 - 38}{5}} = \sqrt{\frac{2}{5}}$$.

So $$r^2 = \frac{2}{5}$$ and $$\alpha + 6r^2 = \frac{48}{5} + 6\cdot\frac{2}{5} = \frac{48 + 12}{5} = \frac{60}{5} = 12$$.

Hence, the correct answer is 12.

Label the unknown lengths as $$PQ=\alpha,\; PR=x,\; QS=y,\; RS=z,\; PS=\beta$$ and the required angle as $$\theta=\angle RPS$$.

Step 1: Fix the angles inside the quadrilateral
Interior angles at the vertices $$Q$$ and $$R$$ are each $$70^\circ$$.
The diagonal $$QS$$ divides $$\angle PQR=70^\circ$$ into $$\angle PQS =15^\circ$$ and $$\angle SQR =70^\circ-15^\circ =55^\circ$$.
Similarly, the diagonal $$PR$$ divides $$\angle QRS=70^\circ$$ into $$\angle PRS =40^\circ$$ and $$\angle QRP =70^\circ-40^\circ =30^\circ$$.

Step 2: Triangle $$\triangle PQR$$
In $$\triangle PQR$$ the three angles are $$\angle PQR =70^\circ,\; \angle QRP =30^\circ,\; \angle QPR =180^\circ-70^\circ-30^\circ =80^\circ.$$
With $$QR=1$$, the sine rule gives $$\frac{\alpha}{\sin30^\circ}=\frac{1}{\sin80^\circ}\;\;\Longrightarrow\;\; \alpha=\frac{\sin30^\circ}{\sin80^\circ}=\frac12\cdot\frac1{\sin80^\circ}.$$

Step 3: Triangle $$\triangle QRS$$
For $$\triangle QRS$$ we already know $$\angle QRS =70^\circ,\; \angle SQR =55^\circ,\; \angle QSR =55^\circ.$$ Hence it is isosceles with $$QR=RS$$, so $$z=RS=QR=1.$$ Applying the sine rule once more gives $$y=QS=\frac{\sin70^\circ}{\sin55^\circ}.$$

Step 4: Triangle $$\triangle PRS$$ and evaluation of $$\sin\theta$$
In $$\triangle PRS$$ the known data are $$PR=x\;(\text{from Step 2}),\; RS=z=1,\; \angle PRS =40^\circ.$$ By the sine rule $$\frac{1}{\sin\theta}=\frac{\beta}{\sin40^\circ}\;\;\Longrightarrow\;\; \sin\theta=\frac{\sin40^\circ}{\beta}.$$

Step 5: Eliminate $$\beta$$ and simplify the required product
Consider the product $$4\alpha\beta\sin\theta=4\alpha\beta\cdot\frac{\sin40^\circ}{\beta}=4\alpha\sin40^\circ.$$ Thus $$\beta$$ and $$\theta$$ disappear - only $$\alpha$$ is still needed.

Step 6: Substitute $$\alpha$$
Using $$\alpha=\dfrac{\sin30^\circ}{\sin80^\circ}$$ obtained in Step 2, $$4\alpha\sin40^\circ =4\left(\frac{\sin30^\circ}{\sin80^\circ}\right)\sin40^\circ =2\cdot\frac{\sin40^\circ}{\sin80^\circ}.$$

Step 7: Recognise the exact value
Since $$\sin80^\circ=2\sin40^\circ\cos40^\circ,$$ $$2\cdot\frac{\sin40^\circ}{\sin80^\circ} =2\cdot\frac{\sin40^\circ}{2\sin40^\circ\cos40^\circ} =\frac1{\cos40^\circ} =\sec40^\circ.$$ Numerically, $$\sec40^\circ\approx1.305.$$

Step 8: Locate the value in the given intervals
$$\sec40^\circ$$ lies between $$1$$ and $$\sqrt2\;( \sqrt2\approx1.414 ).$$
Hence   • it belongs to $$(0,\sqrt{2})$$ (Option A), and
  • it also belongs to $$(1,2)$$ (Option B).

Therefore the correct intervals are:
Option A $$(0,\sqrt{2})$$ and Option B $$(1,2).$$

Circle $$C$$ touches the line $$L: x - 2y = 0$$ at the point $$(2, 1)$$. The equation of a family of circles touching a line $$L = 0$$ at a specific point $$(x_1, y_1)$$ is given by:

$$(x - x_1)^2 + (y - y_1)^2 + \lambda(L) = 0$$

Substituting our values:

$$(x - 2)^2 + (y - 1)^2 + \lambda(x - 2y) = 0$$

$$x^2 - 4x + 4 + y^2 - 2y + 1 + \lambda x - 2\lambda y = 0$$

$$x^2 + y^2 + x(\lambda - 4) - y(2\lambda + 2) + 5 = 0 \quad \text{--- (Equation of Circle } C)$$

The equation of the common chord between Circle $$C$$ and Circle $$C_1: x^2 + y^2 + 2y - 5 = 0$$ is found by subtracting their equations ($$C - C_1 = 0$$):

$$(x^2 + y^2 + x(\lambda - 4) - y(2\lambda + 2) + 5) - (x^2 + y^2 + 2y - 5) = 0$$

$$x(\lambda - 4) - y(2\lambda + 4) + 10 = 0 $$

We are told that $PQ$ is a diameter of circle $$C_1$$. This means the center of circle $$C_1$$ must lie on the line $$PQ$$.

• Center of $$C_1$$: From $$x^2 + y^2 + 2y - 5 = 0$$, the center is $$(0, -1)$$.

Substituting $$(0, -1)$$ into the equation for chord $$PQ$$:

$$0(\lambda - 4) - (-1)(2\lambda + 4) + 10 = 0$$

$$2\lambda + 4 + 10 = 0$$

$$2\lambda = -14 \implies \lambda = -7$$

Substitute $$\lambda = -7$$ back into the equation for Circle $$C$$:

$$x^2 + y^2 + x(-7 - 4) - y(2(-7) + 2) + 5 = 0$$

$$x^2 + y^2 - 11x + 12y + 5 = 0$$

The radius $$r$$ of this circle is calculated using $$r = \sqrt{g^2 + f^2 - c}$$:

• $$g = -\frac{11}{2}$$
• $$f = 6$$
• $$c = 5$$

$$r = \sqrt{\left(-\frac{11}{2}\right)^2 + (6)^2 - 5} = \sqrt{\frac{121}{4} + 36 - 5}$$

$$r = \sqrt{\frac{121 + 144 - 20}{4}} = \sqrt{\frac{245}{4}} = \frac{\sqrt{49 \times 5}}{2} = \frac{7\sqrt{5}}{2}$$

The diameter is $$2r$$:

$$\text{Diameter} = 2 \times \frac{7\sqrt{5}}{2} = 7\sqrt{5}$$

Final Answer:

The diameter of circle $$C$$ is $$7\sqrt{5}$$, which corresponds to Option D.

Let a point on the circle $$x^2 + y^2 = 1$$ be $$(\cos\theta, \sin\theta)$$. The midpoint of the segment from $$(3, 2)$$ to this point is $$\left(\dfrac{3 + \cos\theta}{2},\; \dfrac{2 + \sin\theta}{2}\right)$$.

Let $$h = \dfrac{3 + \cos\theta}{2}$$ and $$k = \dfrac{2 + \sin\theta}{2}$$. Then $$\cos\theta = 2h - 3$$ and $$\sin\theta = 2k - 2$$.

Using $$\cos^2\theta + \sin^2\theta = 1$$, we get $$(2h - 3)^2 + (2k - 2)^2 = 1$$, which simplifies to $$\left(h - \dfrac{3}{2}\right)^2 + (k - 1)^2 = \dfrac{1}{4}$$.

This is a circle with centre $$\left(\dfrac{3}{2}, 1\right)$$ and radius $$r = \dfrac{1}{2}$$.

The given circle is $$x^2 + y^2 + 2x + 4y - 4 = 0$$, which can be rewritten as $$(x+1)^2 + (y+2)^2 = 9$$. This circle has centre $$(-1, -2)$$ and radius $$3$$.

The centres of the required circles lie on this circle, so we parametrise them as $$C = (-1 + 3\cos t,\, -2 + 3\sin t)$$. The radius of each required circle equals $$|CP|$$, where $$P = (-4, 1)$$.

$$|CP|^2 = (-1 + 3\cos t + 4)^2 + (-2 + 3\sin t - 1)^2 = (3 + 3\cos t)^2 + (-3 + 3\sin t)^2$$ $$= 9(1+\cos t)^2 + 9(\sin t - 1)^2 = 9[1 + 2\cos t + \cos^2 t + \sin^2 t - 2\sin t + 1]$$ $$= 9[3 + 2\cos t - 2\sin t].$$

The expression $$2\cos t - 2\sin t$$ has range $$[-2\sqrt{2},\, 2\sqrt{2}]$$, so: $$|CP|^2_{\max} = 9(3 + 2\sqrt{2}), \quad |CP|^2_{\min} = 9(3 - 2\sqrt{2}).$$

Therefore $$r_1 = 3\sqrt{3 + 2\sqrt{2}}$$ and $$r_2 = 3\sqrt{3 - 2\sqrt{2}}$$, giving: $$\frac{r_1}{r_2} = \sqrt{\frac{3 + 2\sqrt{2}}{3 - 2\sqrt{2}}} = \sqrt{\frac{(3+2\sqrt{2})^2}{9 - 8}} = 3 + 2\sqrt{2}.$$

Comparing with $$a + b\sqrt{2}$$, we get $$a = 3$$ and $$b = 2$$, so $$a + b = 5$$.

First, divide the equation of circle $$S$$ by 36 to get standard form: $$x^2 + y^2 - 3x + \frac{10}{3}y + \frac{C}{36} = 0$$.

The centre is $$\left(\frac{3}{2}, -\frac{5}{3}\right)$$ and the radius is $$r = \sqrt{\frac{9}{4} + \frac{25}{9} - \frac{C}{36}} = \sqrt{\frac{81 + 100 - C}{36}} = \frac{\sqrt{181 - C}}{6}$$.

For the circle to exist, we need $$181 - C > 0$$, i.e., $$C < 181$$.

The centre is at $$\left(\frac{3}{2}, -\frac{5}{3}\right)$$, which is in the fourth quadrant.

Condition 1 (no touch/intersection with x-axis): The distance from the centre to the $$x$$-axis is $$\left|-\frac{5}{3}\right| = \frac{5}{3}$$. We need $$r < \frac{5}{3}$$, so $$\frac{\sqrt{181-C}}{6} < \frac{5}{3}$$, giving $$\sqrt{181-C} < 10$$, so $$181 - C < 100$$, i.e., $$C > 81$$.

Condition 2 (no touch/intersection with y-axis): The distance from the centre to the $$y$$-axis is $$\frac{3}{2}$$. We need $$r < \frac{3}{2}$$, so $$\frac{\sqrt{181-C}}{6} < \frac{3}{2}$$, giving $$\sqrt{181-C} < 9$$, so $$181 - C < 81$$, i.e., $$C > 100$$.

The stricter condition is $$C > 100$$.

Now find the intersection of $$x - 2y = 4$$ and $$2x - y = 5$$. From the first equation, $$x = 4 + 2y$$. Substituting: $$2(4 + 2y) - y = 5$$, so $$8 + 3y = 5$$, giving $$y = -1$$ and $$x = 2$$. The point is $$(2, -1)$$.

This point must lie inside the circle: distance from $$(2, -1)$$ to centre $$\left(\frac{3}{2}, -\frac{5}{3}\right)$$ must be less than $$r$$.

Distance squared: $$\left(\frac{1}{2}\right)^2 + \left(\frac{2}{3}\right)^2 = \frac{1}{4} + \frac{4}{9} = \frac{9 + 16}{36} = \frac{25}{36}$$.

We need $$\frac{25}{36} < r^2 = \frac{181-C}{36}$$, so $$25 < 181 - C$$, giving $$C < 156$$.

Combining: $$100 < C < 156$$.

The answer is $$100 < C < 156$$, which is Option D.

The circle is $$x^2 + y^2 + ax + 2ay + c = 0$$ with $$a < 0$$. Comparing with the standard form $$x^2 + y^2 + 2gx + 2fy + c = 0$$, we get $$g = \frac{a}{2}$$ and $$f = a$$. The center is $$\left(-\frac{a}{2}, -a\right)$$ and the radius squared is $$r^2 = g^2 + f^2 - c = \frac{a^2}{4} + a^2 - c = \frac{5a^2}{4} - c$$.

The length of the intercept on the x-axis is given by $$2\sqrt{g^2 - c}$$. Setting this equal to $$2\sqrt{2}$$: $$2\sqrt{\frac{a^2}{4} - c} = 2\sqrt{2}$$, which gives $$\frac{a^2}{4} - c = 2$$ ... (i).

The length of the intercept on the y-axis is given by $$2\sqrt{f^2 - c}$$. Setting this equal to $$2\sqrt{5}$$: $$2\sqrt{a^2 - c} = 2\sqrt{5}$$, which gives $$a^2 - c = 5$$ ... (ii).

Subtracting (i) from (ii): $$a^2 - \frac{a^2}{4} = 3$$, so $$\frac{3a^2}{4} = 3$$, giving $$a^2 = 4$$. Since $$a < 0$$, we have $$a = -2$$.

Substituting back into (i): $$\frac{4}{4} - c = 2$$, so $$c = -1$$.

The center of the circle is $$\left(-\frac{-2}{2}, -(-2)\right) = (1, 2)$$ and $$r^2 = \frac{5 \cdot 4}{4} - (-1) = 5 + 1 = 6$$, so $$r = \sqrt{6}$$.

We need tangent lines perpendicular to $$x + 2y = 0$$, which has slope $$-\frac{1}{2}$$. The perpendicular direction has slope $$2$$. So the tangent lines have the form $$y = 2x + k$$, or equivalently $$2x - y + k = 0$$.

For a tangent to the circle, the distance from the center $$(1, 2)$$ to the line must equal the radius: $$\frac{|2(1) - 2 + k|}{\sqrt{4 + 1}} = \sqrt{6}$$, so $$\frac{|k|}{\sqrt{5}} = \sqrt{6}$$, giving $$|k| = \sqrt{30}$$.

The two tangent lines are $$2x - y + \sqrt{30} = 0$$ and $$2x - y - \sqrt{30} = 0$$. The distance from the origin $$(0, 0)$$ to the line $$2x - y + k = 0$$ is $$\frac{|k|}{\sqrt{5}}$$. For $$k = \sqrt{30}$$, the distance is $$\frac{\sqrt{30}}{\sqrt{5}} = \sqrt{6}$$. For $$k = -\sqrt{30}$$, the distance is also $$\frac{\sqrt{30}}{\sqrt{5}} = \sqrt{6}$$.

Therefore, the shortest distance from the origin to a tangent to this circle perpendicular to $$x + 2y = 0$$ is $$\sqrt{6}$$.

Let $$P$$ be a point on the circle $$(x - 1)^2 + (y - 1)^2 = 1$$. We can parameterize $$P = (1 + \cos\theta,\; 1 + \sin\theta)$$.

Computing $$PA^2$$ where $$A = (1, 4)$$: $$PA^2 = \cos^2\theta + (\sin\theta - 3)^2 = \cos^2\theta + \sin^2\theta - 6\sin\theta + 9 = 10 - 6\sin\theta$$.

Computing $$PB^2$$ where $$B = (1, -5)$$: $$PB^2 = \cos^2\theta + (\sin\theta + 6)^2 = 1 + 12\sin\theta + 36 = 37 + 12\sin\theta$$.

Therefore $$PA^2 + PB^2 = 47 + 6\sin\theta$$. This is maximized when $$\sin\theta = 1$$, i.e., $$\theta = \dfrac{\pi}{2}$$, giving $$P = (1, 2)$$.

Now we check: $$P = (1, 2)$$, $$A = (1, 4)$$, $$B = (1, -5)$$. All three points have the same $$x$$-coordinate equal to 1, so they are collinear and lie on the vertical line $$x = 1$$.

Hence the points $$P$$, $$A$$, and $$B$$ lie on a straight line.

We have $$S_1: x^2 + y^2 = 9$$ with center $$C_1 = (0,0)$$ and radius $$r_1 = 3$$, and $$S_2: (x-2)^2 + y^2 = 1$$ with center $$C_2 = (2,0)$$ and radius $$r_2 = 1$$. Let the variable circle $$S$$ have center $$(h, k)$$ and radius $$R$$.

Since $$S$$ touches $$S_1$$ internally, the distance from $$(h,k)$$ to $$(0,0)$$ satisfies $$\sqrt{h^2 + k^2} = r_1 - R = 3 - R$$. Since $$S$$ touches $$S_2$$ externally, $$\sqrt{(h-2)^2 + k^2} = R + r_2 = R + 1$$.

Adding these two equations gives $$\sqrt{h^2 + k^2} + \sqrt{(h-2)^2 + k^2} = 4$$. This is the equation of an ellipse with foci at $$(0,0)$$ and $$(2,0)$$, where the sum of distances equals $$2a = 4$$, so $$a = 2$$. The distance between foci is $$2c = 2$$, so $$c = 1$$, and $$b^2 = a^2 - c^2 = 4 - 1 = 3$$. The center of the ellipse is $$(1, 0)$$.

The ellipse equation is $$\frac{(x-1)^2}{4} + \frac{y^2}{3} = 1$$. We check each option by substituting. For $$(2, \pm\frac{3}{2})$$: $$\frac{(2-1)^2}{4} + \frac{(3/2)^2}{3} = \frac{1}{4} + \frac{9/4}{3} = \frac{1}{4} + \frac{3}{4} = 1$$. This satisfies the ellipse equation.

The circle is $$x^2 + y^2 - 2x - 4y + 4 = 0$$. Completing the square: $$(x-1)^2 + (y-2)^2 = 1 + 4 - 4 = 1$$. So the centre is $$C = (1, 2)$$ and radius $$r = 1$$.

Two tangents are drawn from an external point $$P$$ to this circle, touching it at points $$A$$ and $$B$$. The angle between the two tangents is $$2\alpha = \tan^{-1}\left(\frac{12}{5}\right)$$. Here $$\alpha$$ denotes the half-angle (the angle each tangent makes with the line $$PC$$).

From the given information, $$\tan(2\alpha) = \frac{12}{5}$$. Using the double-angle formula $$\tan(2\alpha) = \frac{2\tan\alpha}{1 - \tan^2\alpha}$$, we set up the equation: $$\frac{2\tan\alpha}{1 - \tan^2\alpha} = \frac{12}{5}$$.

Cross-multiplying: $$10\tan\alpha = 12(1 - \tan^2\alpha) = 12 - 12\tan^2\alpha$$. Rearranging: $$12\tan^2\alpha + 10\tan\alpha - 12 = 0$$, or equivalently $$6\tan^2\alpha + 5\tan\alpha - 6 = 0$$.

Factoring this quadratic: $$(3\tan\alpha - 2)(2\tan\alpha + 3) = 0$$. This gives $$\tan\alpha = \frac{2}{3}$$ or $$\tan\alpha = -\frac{3}{2}$$. Since $$\alpha \in \left(0, \frac{\pi}{2}\right)$$, we need $$\tan\alpha > 0$$, so $$\tan\alpha = \frac{2}{3}$$.

From $$\tan\alpha = \frac{2}{3}$$, we get $$\sin\alpha = \frac{2}{\sqrt{13}}$$ and $$\cos\alpha = \frac{3}{\sqrt{13}}$$ (using a right triangle with opposite = 2, adjacent = 3, hypotenuse = $$\sqrt{13}$$).

In the right triangle $$\triangle PCA$$ (where the tangent $$PA$$ meets the radius $$CA$$ at a right angle), we have $$\sin\alpha = \frac{CA}{PC} = \frac{r}{PC} = \frac{1}{PC}$$. So $$PC = \frac{1}{\sin\alpha} = \frac{\sqrt{13}}{2}$$.

The tangent length from $$P$$ is $$PA = PB = \sqrt{PC^2 - r^2} = \sqrt{\frac{13}{4} - 1} = \sqrt{\frac{9}{4}} = \frac{3}{2}$$.

Both triangles $$\triangle PAB$$ and $$\triangle CAB$$ share the base $$AB$$. The line $$PC$$ is the axis of symmetry, and it passes through the midpoint $$M$$ of chord $$AB$$, with $$PM \perp AB$$ and $$CM \perp AB$$.

The height from $$P$$ to $$AB$$ is $$PM = PA \cos\alpha = \frac{3}{2} \cdot \frac{3}{\sqrt{13}} = \frac{9}{2\sqrt{13}}$$.

The height from $$C$$ to $$AB$$ is $$CM = CA \cos(\angle ACM)$$. Since $$\angle ACM = \frac{\pi}{2} - \alpha$$ (from the right angle at $$A$$), $$CM = r\cos\left(\frac{\pi}{2} - \alpha\right) = r\sin\alpha = 1 \cdot \frac{2}{\sqrt{13}} = \frac{2}{\sqrt{13}}$$. Alternatively, $$CM = PC - PM = \frac{\sqrt{13}}{2} - \frac{9}{2\sqrt{13}} = \frac{13 - 9}{2\sqrt{13}} = \frac{4}{2\sqrt{13}} = \frac{2}{\sqrt{13}}$$.

The ratio of areas is: $$\frac{\text{Area}(\triangle PAB)}{\text{Area}(\triangle CAB)} = \frac{\frac{1}{2} \cdot AB \cdot PM}{\frac{1}{2} \cdot AB \cdot CM} = \frac{PM}{CM} = \frac{9/(2\sqrt{13})}{2/\sqrt{13}} = \frac{9}{2\sqrt{13}} \times \frac{\sqrt{13}}{2} = \frac{9}{4}$$.

The ratio of areas is $$9 : 4$$, which is Option B.

First, we rewrite the given circle $$x^{2}+y^{2}-2x-6y+6=0$$ in its centre-radius form. To complete the squares we add and subtract the required constants:

$$x^{2}-2x+1+y^{2}-6y+9+6-1-9=0$$ $$\bigl(x-1\bigr)^{2}+\bigl(y-3\bigr)^{2}-4=0$$ $$\bigl(x-1\bigr)^{2}+\bigl(y-3\bigr)^{2}=4.$$

So the centre of the circle is $$C(1,3)$$ and the radius is $$r=2.$$

The external point from which the tangents are drawn is $$P(-1,1).$$ Using the distance formula, the distance of $$P$$ from the centre $$C$$ is

$$CP=\sqrt{(1-(-1))^{2}+(3-1)^{2}}=\sqrt{2^{2}+2^{2}}=\sqrt{8}=2\sqrt{2}.$$

The length of each tangent drawn from an external point to a circle is given by the formula $$\text{(tangent length)}=\sqrt{CP^{2}-r^{2}}.$$ Substituting $$CP=2\sqrt{2}$$ and $$r=2$$ we obtain

$$PA=PB=\sqrt{(2\sqrt{2})^{2}-2^{2}}=\sqrt{8-4}=\sqrt{4}=2.$$

Let $$A(x_{1},y_{1})$$ and $$B(x_{2},y_{2})$$ be the points of contact. For the circle $$\bigl(x-1\bigr)^{2}+\bigl(y-3\bigr)^{2}=4,$$ the tangent at a point $$(x_{1},y_{1})$$ on the circle is $$(x_{1}-1)(x-1)+(y_{1}-3)(y-3)=4.$$ Since this tangent passes through $$P(-1,1),$$ we substitute:

$$(x_{1}-1)(-1-1)+(y_{1}-3)(1-3)=4,$$ $$-2\bigl[(x_{1}-1)+(y_{1}-3)\bigr]=4,$$ $$(x_{1}-1)+(y_{1}-3)=-2,$$ $$x_{1}+y_{1}=2.$$

Along with the fact that $$A$$ lies on the circle, i.e. $$(x_{1}-1)^{2}+(y_{1}-3)^{2}=4,$$ we solve the system

$$\begin{cases} x_{1}+y_{1}=2,\\ (x_{1}-1)^{2}+(y_{1}-3)^{2}=4. \end{cases}$$

Putting $$y_{1}=2-x_{1}$$ in the second equation,

$$(x_{1}-1)^{2}+(2-x_{1}-3)^{2}=4,$$ $$(x_{1}-1)^{2}+(-1-x_{1})^{2}=4,$$ $$x_{1}^{2}-2x_{1}+1+x_{1}^{2}+2x_{1}+1=4,$$ $$2x_{1}^{2}+2=4,$$ $$x_{1}^{2}=1,$$ $$x_{1}=1\quad\text{or}\quad x_{1}=-1.$$

Corresponding $$y_{1}=2-x_{1}$$ gives

$$A(1,1)\quad\text{and}\quad B(-1,3).$$

The length of chord $$AB$$ is therefore

$$AB=\sqrt{(-1-1)^{2}+(3-1)^{2}}=\sqrt{(-2)^{2}+2^{2}}=\sqrt{8}=2\sqrt{2}.$$

Now we are told that a point $$D$$ on the circle satisfies $$AD=AB=2\sqrt{2}.$$ Let $$D(x,y)$$ be such a point. We require

$$\bigl(x-1\bigr)^{2}+\bigl(y-3\bigr)^{2}=4\quad\text{(because }D\text{ is on the circle)},$$ $$\bigl(x-1\bigr)^{2}+\bigl(y-1\bigr)^{2}=8\quad\text{(because }AD=2\sqrt{2}).$$

Subtracting the first equation from the second eliminates the $$\bigl(x-1\bigr)^{2}$$ term:

$$(y-1)^{2}-(y-3)^{2}=8-4,$$ $$y^{2}-2y+1-\bigl(y^{2}-6y+9\bigr)=4,$$ $$y^{2}-2y+1-y^{2}+6y-9=4,$$ $$4y-8=4,$$ $$4y=12,$$ $$y=3.$$

Substituting $$y=3$$ in the circle equation,

$$(x-1)^{2}=4,$$ $$x-1=\pm2,$$ $$x=3\quad\text{or}\quad x=-1.$$

The point $$(-1,3)$$ is already the point $$B,$$ so the new point is $$D(3,3).$$

Thus the vertices of $$\triangle ABD$$ are $$A(1,1),\;B(-1,3),\;D(3,3).$$

Using the coordinate (shoelace) formula, the area of a triangle with vertices $$(x_{1},y_{1}),\,(x_{2},y_{2}),\,(x_{3},y_{3})$$ is

$$\text{Area}=\frac12\Bigl|x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1})+x_{3}(y_{1}-y_{2})\Bigr|.$$

Substituting $$A(1,1),\;B(-1,3),\;D(3,3)$$ we get

$$\text{Area}=\frac12\Bigl[\,1(3-3)+(-1)(3-1)+3(1-3)\Bigr]$$ $$=\frac12\bigl[\,0-2-6\bigr]=\frac12(-8)=4.$$

The absolute value has been taken in the last step. Therefore the area of $$\triangle ABD$$ is $$4.$$

Hence, the correct answer is Option C.

We are told that the circle has centre $$C(2,\,3)$$ and passes through the origin $$O(0,\,0)$$. Using the distance formula, the radius $$r$$ is

$$r = OC = \sqrt{(2 - 0)^2 + (3 - 0)^2} = \sqrt{4 + 9} = \sqrt{13}.$$

Any point $$P(x,\,y)$$ lying on this circle must therefore satisfy the standard equation

$$CP = r \;\Longrightarrow\; (x - 2)^2 + (y - 3)^2 = 13.$$

We are further told that the line segment $$OC$$ is perpendicular to the line segments $$CP$$ and $$CQ$$. A perpendicularity condition in vector form is

$$\vec{CP}\cdot\vec{CO}=0.$$

Here, $$\vec{CO}$$ is the vector from $$C$$ to $$O$$, so

$$\vec{CO}=O-C=(0-2,\;0-3)=(-2,\,-3).$$

If $$P(x,\,y)$$ is any required point, then $$\vec{CP}=P-C=(x-2,\;y-3).$$ The dot-product condition becomes

$$\bigl(x-2,\;y-3\bigr)\cdot(-2,\,-3)=0.$$

Evaluating the dot product, we get

$$-2(x-2)\;-\;3(y-3)=0 \;\Longrightarrow\; -2x+4-3y+9=0 \;\Longrightarrow\; -2x-3y+13=0.$$

Rearranging, the equation of the line along which both $$P$$ and $$Q$$ must lie is therefore

$$2x + 3y = 13.$$

Because this line is perpendicular to $$OC$$ and passes through the centre $$C(2,\,3)$$, every admissible point must satisfy both the circle equation and this linear equation. To remove one variable, we express $$y$$ from the linear equation:

$$3y = 13 - 2x \;\Longrightarrow\; y = 3 - \dfrac{2}{3}(x-2).$$

Substituting this expression for $$y$$ into the circle equation $$ (x-2)^2 + (y-3)^2 = 13$$, we write

$$ (x-2)^2 + \left[\,3 - \frac{2}{3}(x-2) - 3\right]^2 = 13.$$

The square bracket simplifies because $$3 - 3=0$$, leaving

$$ (x-2)^2 + \left[-\frac{2}{3}(x-2)\right]^2 = 13.$$

First compute each term: $$ (x-2)^2 = (x-2)^2, \qquad \left[-\frac{2}{3}(x-2)\right]^2 = \frac{4}{9}(x-2)^2. $$ Adding them, $$ (x-2)^2 + \frac{4}{9}(x-2)^2 = \left(1+\frac{4}{9}\right)(x-2)^2 = \frac{13}{9}(x-2)^2. $$ Therefore $$ \frac{13}{9}(x-2)^2 = 13. $$ Multiplying both sides by $$\dfrac{9}{13}$$, we obtain $$ (x-2)^2 = 9. $$ Taking square roots, $$ x-2 = \pm 3 \;\Longrightarrow\; x = 2 \pm 3. $$ Hence the two possible $$x$$-coordinates are $$ x_1 = 5, \qquad x_2 = -1. $$

We now find the corresponding $$y$$-coordinates from $$y = 3 - \dfrac{2}{3}(x-2)$$.

For $$x = 5$$: $$ y = 3 - \frac{2}{3}(5-2) = 3 - \frac{2}{3}\times 3 = 3 - 2 = 1, $$ so one point is $$P(5,\,1).$$

For $$x = -1$$: $$ y = 3 - \frac{2}{3}(-1-2) = 3 - \frac{2}{3}\times(-3) = 3 + 2 = 5, $$ so the other point is $$Q(-1,\,5).$$

Thus the required unordered pair of points is

$$\{P,\,Q\} = \{(-1,\,5),\,(5,\,1)\}.$$

Comparing with the given options, this matches Option D.

Hence, the correct answer is Option D.

The tangent to the circle $$x^2 + y^2 = 25$$ at the point $$R(3, 4)$$ has the equation $$3x + 4y = 25$$.

To find where this meets the $$x$$-axis, set $$y = 0$$: $$3x = 25$$, so $$P = \left(\frac{25}{3}, 0\right)$$.

To find where it meets the $$y$$-axis, set $$x = 0$$: $$4y = 25$$, so $$Q = \left(0, \frac{25}{4}\right)$$.

Now we have the triangle $$OPQ$$ with vertices $$O = (0, 0)$$, $$P = \left(\frac{25}{3}, 0\right)$$, $$Q = \left(0, \frac{25}{4}\right)$$.

The side lengths are: $$OP = \frac{25}{3}$$, $$OQ = \frac{25}{4}$$, and $$PQ = \sqrt{\left(\frac{25}{3}\right)^2 + \left(\frac{25}{4}\right)^2} = 25\sqrt{\frac{1}{9} + \frac{1}{16}} = 25\sqrt{\frac{25}{144}} = \frac{25 \times 5}{12} = \frac{125}{12}$$.

For the incentre formula, we label: side $$a = PQ = \frac{125}{12}$$ (opposite vertex $$O$$), side $$b = OQ = \frac{25}{4}$$ (opposite vertex $$P$$), side $$c = OP = \frac{25}{3}$$ (opposite vertex $$Q$$).

The incentre is $$I = \frac{a \cdot O + b \cdot P + c \cdot Q}{a + b + c}$$.

First, $$a + b + c = \frac{125}{12} + \frac{25}{4} + \frac{25}{3} = \frac{125}{12} + \frac{75}{12} + \frac{100}{12} = \frac{300}{12} = 25$$.

$$I_x = \frac{a \cdot 0 + b \cdot \frac{25}{3} + c \cdot 0}{25} = \frac{\frac{25}{4} \cdot \frac{25}{3}}{25} = \frac{625}{300} = \frac{25}{12}$$.

$$I_y = \frac{a \cdot 0 + b \cdot 0 + c \cdot \frac{25}{4}}{25} = \frac{\frac{25}{3} \cdot \frac{25}{4}}{25} = \frac{625}{300} = \frac{25}{12}$$.

So the incentre is $$I = \left(\frac{25}{12}, \frac{25}{12}\right)$$.

The circle passes through the origin $$O(0,0)$$ and has its centre at $$I$$. So the radius $$r$$ equals the distance from $$I$$ to $$O$$.

$$r^2 = \left(\frac{25}{12}\right)^2 + \left(\frac{25}{12}\right)^2 = 2 \cdot \frac{625}{144} = \frac{1250}{144} = \frac{625}{72}$$.

The answer is $$\frac{625}{72}$$, which is Option C.

The tangent to the circle at point $$(2, 5)$$ is $$2x - y + 1 = 0$$. The centre of the circle lies on the line $$x - 2y = 4$$.

The radius at the point of tangency is perpendicular to the tangent line. The tangent $$2x - y + 1 = 0$$ has slope $$2$$, so the radius through $$(2, 5)$$ has slope $$-\frac{1}{2}$$.

The equation of the line through $$(2, 5)$$ with slope $$-\frac{1}{2}$$ is: $$y - 5 = -\frac{1}{2}(x - 2)$$, which simplifies to $$x + 2y = 12$$ $$-(1)$$.

The centre lies on both this line and $$x - 2y = 4$$ $$-(2)$$.

Adding $$(1)$$ and $$(2)$$: $$2x = 16$$, so $$x = 8$$. From $$(2)$$: $$8 - 2y = 4$$, so $$y = 2$$. The centre is $$(8, 2)$$.

The radius is the distance from the centre $$(8, 2)$$ to the point of tangency $$(2, 5)$$: $$r = \sqrt{(8-2)^2 + (2-5)^2} = \sqrt{36 + 9} = \sqrt{45} = 3\sqrt{5}$$.

This matches Option A: $$3\sqrt{5}$$.

The two circles are $$x^2 + y^2 - 10x - 10y + 41 = 0$$ and $$x^2 + y^2 - 16x - 10y + 80 = 0$$.

For the first circle, completing the square: $$(x-5)^2 + (y-5)^2 = 25 + 25 - 41 = 9$$. So centre $$C_1 = (5, 5)$$ and radius $$r_1 = 3$$.

For the second circle: $$(x-8)^2 + (y-5)^2 = 64 + 25 - 80 = 9$$. So centre $$C_2 = (8, 5)$$ and radius $$r_2 = 3$$.

The distance between centres is $$d = \sqrt{(8-5)^2 + (5-5)^2} = 3$$.

Now let us check each statement:

Option A: "Distance between centres is the average of radii." The average of $$r_1$$ and $$r_2$$ is $$\frac{3+3}{2} = 3 = d$$. This is TRUE.

Option B: "Both circles' centres lie inside the region of one another." For $$C_2 = (8,5)$$ to be inside circle 1, we need $$(8-5)^2 + (5-5)^2 = 9 < r_1^2 = 9$$. Since $$9$$ is not strictly less than $$9$$, $$C_2$$ lies ON circle 1, not inside it. Similarly $$C_1$$ lies ON circle 2. So this statement is INCORRECT.

Option C: "Both circles pass through the centre of each other." Substituting $$C_2 = (8,5)$$ into circle 1: $$(8-5)^2 + (5-5)^2 = 9 = r_1^2$$. Yes, $$C_2$$ lies on circle 1. Similarly $$(5-8)^2 + (5-5)^2 = 9 = r_2^2$$, so $$C_1$$ lies on circle 2. This is TRUE.

Option D: "Circles have two intersection points." Since $$|r_1 - r_2| < d < r_1 + r_2$$ becomes $$0 < 3 < 6$$, which is true, the circles intersect at two points. This is TRUE.

The incorrect statement is Option B, since the centres lie ON the other circle, not inside it.

The circle is tangent to the $$y$$-axis at the point $$(0,6)$$. Whenever a circle touches a line, the radius drawn to the point of contact is perpendicular to that line. Here the tangent line is the $$y$$-axis, whose equation is $$x=0$$. Hence the centre of the circle must lie exactly one radius away from this line, on the horizontal through $$(0,6)$$. So if we denote the radius by $$r$$, the centre is $$(r,\,6)$$. (The $$x$$-coordinate is $$r$$ units to the right of the $$y$$-axis, while the $$y$$-coordinate remains $$6$$ to stay level with the point of contact.)

With centre $$(r,6)$$ and radius $$r$$, the equation of the circle is

$$ (x-r)^2 + (y-6)^2 = r^2. $$

Now we use the information about the intercept on the $$x$$-axis. The $$x$$-axis is the line $$y = 0$$. To find the points where the circle meets this axis we substitute $$y = 0$$ in the equation:

$$ (x - r)^2 + (0 - 6)^2 = r^2. $$

Simplifying the constant term, we get

$$ (x - r)^2 + 36 = r^2. $$

Rearranging,

$$ (x - r)^2 = r^2 - 36. $$

Taking square roots gives the two intersection $$x$$-coordinates:

$$ x = r \pm \sqrt{\,r^2 - 36\,}. $$

Thus the two points of intersection on the $$x$$-axis are

$$\bigl(r - \sqrt{r^2 - 36},\,0\bigr) \quad\text{and}\quad \bigl(r + \sqrt{r^2 - 36},\,0\bigr).$$

The length of the intercept cut off on the $$x$$-axis is the horizontal distance between these two points. Subtracting their $$x$$-coordinates,

$$ \text{Intercept length} = \bigl[r + \sqrt{r^2 - 36}\bigr] - \bigl[r - \sqrt{r^2 - 36}\bigr] = 2\sqrt{\,r^2 - 36\,}. $$

We are told that this length equals $$6\sqrt{5}$$, so

$$ 2\sqrt{\,r^2 - 36\,} = 6\sqrt{5}. $$

Dividing both sides by $$2$$, we have

$$ \sqrt{\,r^2 - 36\,} = 3\sqrt{5}. $$

Now we square both sides (since both sides are non-negative):

$$ r^2 - 36 = \bigl(3\sqrt{5}\bigr)^2 = 9 \times 5 = 45. $$

Adding $$36$$ to both sides gives

$$ r^2 = 45 + 36 = 81. $$

Finally, taking the positive square root (radius is always positive),

$$ r = 9. $$

Hence, the correct answer is Option B.

First we rewrite the circle equation $$4x^2 + 4y^2 + 120x + 675 = 0$$ in its standard form. Dividing every term by $$4$$ gives $$x^2 + y^2 + 30x + \tfrac{675}{4} = 0.$$ To complete the square for the $$x$$-terms we use $$(x + 15)^2 = x^2 + 30x + 225.$$ Substituting this, we obtain $$(x + 15)^2 - 225 + y^2 + \tfrac{675}{4} = 0.$$ Combining the constants, $$-225 + \tfrac{675}{4} = -\tfrac{225}{1} + \tfrac{675}{4} = -\tfrac{900}{4} + \tfrac{675}{4} = -\tfrac{225}{4}.$$ Hence $$(x + 15)^2 + y^2 = \tfrac{225}{4}.$$ So the centre of the circle is $$(-15,\,0)$$ and the radius is $$R = \sqrt{\tfrac{225}{4}} = \tfrac{15}{2}.$$

The parabola is $$y^2 = 30x,$$ which can be compared with $$y^2 = 4ax.$$ Here $$4a = 30 \;\Rightarrow\; a = 7.5.$$

For the parabola $$y^2 = 4ax,$$ the standard tangent at a point $$(x_1,\,y_1)$$ on it is $$y\,y_1 = 2a\,(x + x_1).$$ Using $$2a = 15,$$ the tangent becomes $$y\,y_1 = 15\,(x + x_1).$$

The required tangent must also pass through the given point $$(-30,\,0).$$ Substituting $$x = -30,\; y = 0$$ into the tangent equation gives $$0 \cdot y_1 = 15\,(-30 + x_1) \;\Longrightarrow\; -30 + x_1 = 0 \;\Longrightarrow\; x_1 = 30.$$

Since $$(x_1,\,y_1)$$ lies on the parabola, we have $$y_1^2 = 30x_1 = 30 \times 30 = 900,$$ so $$y_1 = \pm 30.$$ Thus there are two tangents through $$(-30,0):$$

1. When $$y_1 = 30:$$ $$30y = 15(x + 30) \;\Rightarrow\; 2y = x + 30 \;\Rightarrow\; x - 2y + 30 = 0.$$

2. When $$y_1 = -30:$$ $$-30y = 15(x + 30) \;\Rightarrow\; -2y = x + 30 \;\Rightarrow\; x + 2y + 30 = 0.$$

Each of these lines cuts the circle in a chord. For a circle, the length of the chord cut by a line is given by $$\text{Chord length} = 2\sqrt{R^2 - d^2},$$ where $$d$$ is the perpendicular distance from the centre to the line.

We calculate this distance for one of the lines; both will give the same value because they are symmetric.

Taking $$x - 2y + 30 = 0,$$ the distance from the centre $$(-15,0)$$ is $$d = \frac{|\,(-15) - 2(0) + 30\,|}{\sqrt{1^2 + (-2)^2}} = \frac{|15|}{\sqrt{1 + 4}} = \frac{15}{\sqrt{5}}.$$

Now we evaluate $$R^2 - d^2 = \left(\tfrac{15}{2}\right)^2 - \left(\tfrac{15}{\sqrt{5}}\right)^2 = \tfrac{225}{4} - \tfrac{225}{5} = \tfrac{225}{4} - \tfrac{180}{4} = \tfrac{45}{4}.$$

Therefore the chord length is $$2\sqrt{\tfrac{45}{4}} = 2 \times \frac{\sqrt{45}}{2} = \sqrt{45} = 3\sqrt{5}.$$

Hence, the correct answer is Option C.

We begin by rewriting each of the three given sets in centre-radius form, because that will allow us to compare their positions in the plane very easily.

For the set $$A$$ we have the equation

$$2x^{2}+2y^{2}-2x-2y=1.$$

Dividing by $$2$$ gives

$$x^{2}+y^{2}-x-y=\frac12.$$

Now we complete the square separately in $$x$$ and $$y$$. The standard identity to remember is

$$u^{2}-2au=(u-a)^{2}-a^{2},$$

so subtracting a first-degree term and then adding (\dfrac14$$ makes a perfect square. Applying this, $$x^{2}-x=\Bigl(x-\tfrac12\Bigr)^{2}-\Bigl(\tfrac12\Bigr)^{2},\qquad y^{2}-y=\Bigl(y-\tfrac12\Bigr)^{2}-\Bigl(\tfrac12\Bigr)^{2}.$$ Therefore $$x^{2}+y^{2}-x-y =\Bigl(x-\tfrac12\Bigr)^{2}+\Bigl(y-\tfrac12\Bigr)^{2} -\Bigl(\tfrac12\Bigr)^{2}-\Bigl(\tfrac12\Bigr)^{2} =\Bigl(x-\tfrac12\Bigr)^{2}+\Bigl(y-\tfrac12\Bigr)^{2}-\frac12.$$ Setting this equal to (\dfrac12$$ gives

$$\Bigl(x-\tfrac12\Bigr)^{2}+\Bigl(y-\tfrac12\Bigr)^{2}=1.$$

Hence $$A$$ is the circle with centre $$\bigl(\tfrac12,\tfrac12\bigr)$$ and radius $$1.$$

For the set $$B$$ we start from

$$4x^{2}+4y^{2}-16y+7=0.$$

Dividing by $$4$$ gives

$$x^{2}+y^{2}-4y+\frac74=0.$$

We complete the square in $$y$$ only (no first-degree term in $$x$$):

$$y^{2}-4y=\bigl(y-2\bigr)^{2}-4.$$

Substituting this,

$$x^{2}+\bigl(y-2\bigr)^{2}-4+\frac74=0 \;\Longrightarrow\; x^{2}+\bigl(y-2\bigr)^{2}=4-\frac74=\frac{16-7}{4}=\frac94.$$

So $$B$$ is the circle with centre $$(0,\,2)$$ and radius $$\sqrt{\tfrac94}=\tfrac32.$$

For the set $$C$$ the inequality is

$$x^{2}+y^{2}-4x-2y+5\le r^{2}.$$

Again we complete the squares. First, for $$x$$:

$$x^{2}-4x=(x-2)^{2}-4,$$

and for $$y$$:

$$y^{2}-2y=(y-1)^{2}-1.$$

Hence

$$x^{2}+y^{2}-4x-2y+5 =(x-2)^{2}-4+(y-1)^{2}-1+5 =(x-2)^{2}+(y-1)^{2}.$$

Thus

$$C=\Bigl\{(x,y)\in\mathbb R^{2}\mid (x-2)^{2}+(y-1)^{2}\le r^{2}\Bigr\},$$

which is the closed disc with centre $$(2,\,1)$$ and radius $$|r|.$$ (Because a radius is non-negative, we may as well write simply $$r\ge0$$.)

Our task is to choose the smallest possible $$r$$ so that every point of $$A\cup B$$ lies inside this disc. Geometrically, that means

$$r=\max\bigl\{\text{distance from }(2,1)\text{ to any point of }A\cup B\bigr\}.$$

For each of the two circles, the farthest point from a fixed external point is obtained by moving from the circle’s centre directly away from the fixed point. Therefore, for a circle with centre $$O_{i}$$ and radius $$\rho_{i},$$ the maximum distance to $$(2,1)$$ is

$$\bigl|O_{i}(2,1)\bigr|+\rho_{i}.$$

We now compute these quantities one by one.

Circle A. Centre $$O_{A}=\bigl(\tfrac12,\tfrac12\bigr).$$ The distance between $$(2,1)$$ and $$O_{A}$$ is

$$\sqrt{(2-\tfrac12)^{2}+(1-\tfrac12)^{2}} =\sqrt{\bigl(\tfrac32\bigr)^{2}+\bigl(\tfrac12\bigr)^{2}} =\sqrt{\frac94+\frac14} =\sqrt{\frac{10}{4}} =\frac{\sqrt{10}}{2}.$$

The radius of circle $$A$$ is $$1,$$ so the farthest point of $$A$$ from $$(2,1)$$ is at a distance

$$d_{A}^{\max}=\frac{\sqrt{10}}{2}+1.$$

Circle B. Centre $$O_{B}=(0,2).$$ The distance between $$(2,1)$$ and $$O_{B}$$ is

$$\sqrt{(2-0)^{2}+(1-2)^{2}}=\sqrt{4+1}=\sqrt5.$$

The radius of circle $$B$$ is $$\tfrac32,$$ hence the farthest point of $$B$$ from $$(2,1)$$ is at a distance

$$d_{B}^{\max}=\sqrt5+\frac32.$$

Between these two numbers, the larger one determines the required radius. Observe that

$$\sqrt5+\frac32 \;-\;\Bigl(\,\frac{\sqrt{10}}{2}+1\Bigr) =\sqrt5+\frac32-\frac{\sqrt{10}}{2}-1 =\Bigl(\sqrt5-\frac{\sqrt{10}}{2}\Bigr)+\frac12 >0,$$

so indeed $$d_{B}^{\max}$$ is the greater. Consequently the minimum value of $$r$$ that works is

$$r_{\min}=d_{B}^{\max}=\sqrt5+\frac32=\frac{3+2\sqrt5}{2}.$$

Because $$r\ge0,$$ we also have $$|r|=r,$$ so

$$\min|r|=\frac{3+2\sqrt5}{2}.$$

Hence, the correct answer is Option C.

The first circle is $$x^2 + y^2 - 2x - 6y + 6 = 0$$. We rewrite it in standard form by completing the square: $$(x - 1)^2 + (y - 3)^2 = 1 + 9 - 6 = 4$$.

So the center of the first circle is $$C_1 = (1, 3)$$ and its radius is $$r_1 = 2$$.

A diameter of this circle is a chord that passes through $$C_1 = (1, 3)$$ and has length $$2r_1 = 4$$, so the half-chord length is $$\ell = 2$$.

This diameter is a chord of the second circle $$C$$ whose center is $$C_2 = (2, 1)$$. Let the radius of circle $$C$$ be $$R$$.

We compute the distance from the center of $$C$$ to the center of the first circle: $$d = \sqrt{(2 - 1)^2 + (1 - 3)^2} = \sqrt{1 + 4} = \sqrt{5}$$.

Since the diameter of the first circle passes through $$C_1 = (1, 3)$$, the perpendicular from $$C_2$$ to this chord meets it at $$C_1$$. This is because $$C_1$$ is the midpoint of every diameter of the first circle, and the perpendicular from the center of a circle to a chord bisects the chord. Here, the perpendicular distance from $$C_2$$ to the chord equals the distance from $$C_2$$ to the midpoint $$C_1$$ only when $$C_2$$, $$C_1$$, and the foot of the perpendicular are configured appropriately.

Actually, for any chord of circle $$C$$, if the perpendicular from $$C_2$$ to the chord has length $$d_\perp$$, then $$R^2 = d_\perp^2 + \ell^2$$. The midpoint of the chord (which is a diameter of the first circle) is $$C_1 = (1, 3)$$. The perpendicular from $$C_2$$ to the chord passes through the midpoint of the chord. So the perpendicular distance equals the distance from $$C_2$$ to the midpoint $$C_1$$, which is $$d = \sqrt{5}$$.

Applying the relation $$R^2 = d^2 + \ell^2 = 5 + 4 = 9$$, we get $$R = 3$$.

Hence, the answer is $$3$$.

The two circles are $$x^2 + y^2 - 10x - 10y + 41 = 0$$ and $$x^2 + y^2 - 24x - 10y + 160 = 0$$.

First circle: $$(x-5)^2 + (y-5)^2 = 25 + 25 - 41 = 9$$. Center $$C_1 = (5, 5)$$, radius $$r_1 = 3$$.

Second circle: $$(x-12)^2 + (y-5)^2 = 144 + 25 - 160 = 9$$. Center $$C_2 = (12, 5)$$, radius $$r_2 = 3$$.

Distance between centers: $$d = \sqrt{(12-5)^2 + (5-5)^2} = 7$$.

Since $$d = 7 > r_1 + r_2 = 6$$, the circles are external to each other. The minimum distance between points on the two circles is $$d - r_1 - r_2 = 7 - 3 - 3 = 1$$.

The minimum distance is 1.

The tangent to the parabola $$y^2 = 4x$$ at the point $$(t^2, 2t)$$ is given by $$ty = x + t^2$$, which can be written as $$x - ty + t^2 = 0$$.

For this line to be tangent to the circle $$(x-3)^2 + y^2 = 9$$ (center $$(3, 0)$$, radius $$3$$), the perpendicular distance from $$(3, 0)$$ to the line must equal 3: $$\frac{|3 - 0 + t^2|}{\sqrt{1 + t^2}} = 3$$.

Since $$t^2 + 3 > 0$$, we square both sides: $$(3 + t^2)^2 = 9(1 + t^2)$$. Expanding: $$9 + 6t^2 + t^4 = 9 + 9t^2$$, which gives $$t^4 - 3t^2 = 0$$, so $$t^2(t^2 - 3) = 0$$. Thus $$t^2 = 3$$, and for the first quadrant point on the parabola, $$t = \sqrt{3}$$.

The tangent line is $$\sqrt{3}\,y = x + 3$$. The point of contact on the parabola is $$(t^2, 2t) = (3, 2\sqrt{3})$$, so $$c = 3$$ and $$d = 2\sqrt{3}$$.

For the point of contact on the circle, we find the foot of the perpendicular from the center $$(3, 0)$$ to the line $$x - \sqrt{3}\,y + 3 = 0$$. Using the formula: $$(a, b) = \left(3 - \frac{3 + 0 + 3}{1 + 3} \cdot 1,\; 0 - \frac{3 + 0 + 3}{1 + 3} \cdot (-\sqrt{3})\right) = \left(3 - \frac{3}{2},\; \frac{3\sqrt{3}}{2}\right) = \left(\frac{3}{2},\; \frac{3\sqrt{3}}{2}\right)$$.

So $$a = \dfrac{3}{2}$$. Therefore, $$2(a + c) = 2\left(\dfrac{3}{2} + 3\right) = 2 \times \dfrac{9}{2} = 9$$.

Let the square $$ABCD$$ have vertices $$A = (0, 0)$$, $$B = (1, 0)$$, $$C = (1, 1)$$, $$D = (0, 1)$$. Circle $$C_1$$ is centered at $$A = (0,0)$$ with radius 1.

Circle $$C_2$$ touches lines $$AB$$ (the $$x$$-axis) and $$AD$$ (the $$y$$-axis), so its center is at $$(r, r)$$ for some radius $$r > 0$$. Since $$C_2$$ touches $$C_1$$ internally ($$C_2$$ is inside the region bounded by $$C_1$$), the distance between centers equals $$1 - r$$: $$r\sqrt{2} = 1 - r$$, giving $$r(\sqrt{2} + 1) = 1$$, so $$r = \frac{1}{\sqrt{2} + 1} = \sqrt{2} - 1$$.

So $$C_2$$ has center $$(\sqrt{2} - 1, \sqrt{2} - 1)$$ and radius $$r = \sqrt{2} - 1$$.

A tangent from $$C = (1, 1)$$ to circle $$C_2$$ meets $$AB$$ (the segment from $$(0,0)$$ to $$(1,0)$$, i.e., the $$x$$-axis with $$0 \leq x \leq 1$$) at point $$E$$. The tangent line passes through $$C = (1,1)$$ and touches $$C_2$$. Let $$E = (e, 0)$$ on $$AB$$.

The line through $$(1, 1)$$ and $$(e, 0)$$ has equation: $$y - 0 = \frac{1 - 0}{1 - e}(x - e)$$, i.e., $$x - (1-e)y - e = 0$$.

The distance from center $$(\sqrt{2}-1, \sqrt{2}-1)$$ to this line equals $$r = \sqrt{2} - 1$$:

$$\frac{|(\sqrt{2}-1) - (1-e)(\sqrt{2}-1) - e|}{\sqrt{1 + (1-e)^2}} = \sqrt{2} - 1$$.

The numerator simplifies: $$(\sqrt{2}-1)[1 - (1-e)] - e = (\sqrt{2}-1)e - e = e(\sqrt{2} - 2)$$. Taking absolute value: $$e(2 - \sqrt{2})$$.

So $$\frac{e(2 - \sqrt{2})}{\sqrt{1 + (1-e)^2}} = \sqrt{2} - 1$$.

Squaring both sides: $$\frac{e^2(2 - \sqrt{2})^2}{1 + (1-e)^2} = (\sqrt{2} - 1)^2 = 3 - 2\sqrt{2}$$.

Note $$(2 - \sqrt{2})^2 = 6 - 4\sqrt{2} = 2(3 - 2\sqrt{2})$$. So $$\frac{2e^2(3 - 2\sqrt{2})}{1 + (1-e)^2} = 3 - 2\sqrt{2}$$, giving $$\frac{2e^2}{1 + (1-e)^2} = 1$$.

Therefore $$2e^2 = 1 + 1 - 2e + e^2 = 2 - 2e + e^2$$, so $$e^2 + 2e - 2 = 0$$. Using the quadratic formula: $$e = \frac{-2 + \sqrt{4 + 8}}{2} = \frac{-2 + 2\sqrt{3}}{2} = -1 + \sqrt{3}$$.

Then $$EB = 1 - e = 1 - (-1 + \sqrt{3}) = 2 - \sqrt{3}$$. So $$\alpha + \sqrt{3}\beta = 2 - \sqrt{3}$$, giving $$\alpha = 2$$ and $$\beta = -1$$.

Therefore $$\alpha + \beta = 2 + (-1) = 1$$.

We start with the given family of circles

$$x^{2}+y^{2}+px+(1-p)y+5=0.$$

The standard form of the general circle is

$$x^{2}+y^{2}+2gx+2fy+c=0,$$

whose centre is $$(-g,\,-f)$$ and whose radius is $$R=\sqrt{g^{2}+f^{2}-c}\,.$$

Comparing coefficients we obtain

$$2g=p \;\;\Longrightarrow\;\; g=\dfrac{p}{2},$$

$$2f=1-p \;\;\Longrightarrow\;\; f=\dfrac{1-p}{2},$$

and

$$c=5.$$

Substituting $$g,\;f,\;c$$ in the radius formula gives

$$R=\sqrt{\left(\dfrac{p}{2}\right)^{2}+\left(\dfrac{1-p}{2}\right)^{2}-5} =\sqrt{\dfrac{p^{2}}{4}+\dfrac{1-2p+p^{2}}{4}-5}.$$

Simplifying the expression under the square root, we have

$$R=\sqrt{\dfrac{2p^{2}-2p-19}{4}} =\dfrac{1}{2}\sqrt{\,2p^{2}-2p-19\,}.$$

The problem states that the radius varies in the interval $$(0,\,5],$$ therefore

$$0\lt R\le 5 \;\;\Longrightarrow\;\; 0\lt \dfrac{1}{2}\sqrt{\,2p^{2}-2p-19\,}\le 5.$$

Squaring and multiplying by $$4$$ eliminates the square root and the denominator:

$$0\lt 2p^{2}-2p-19\le 100.$$

Thus the two simultaneous inequalities to be satisfied by $$p$$ are

$$\text{(i)}\; 2p^{2}-2p-19 \gt 0,$$

$$\text{(ii)}\; 2p^{2}-2p-119 \le 0.$$

First, solve (i):

$$2p^{2}-2p-19=0 \;\;\Longrightarrow\;\; p=\dfrac{2\pm\sqrt{(-2)^{2}-4\cdot2\cdot(-19)}}{4} =\dfrac{2\pm2\sqrt{39}}{4} =\dfrac{1\pm\sqrt{39}}{2}.$$

Numerically,

$$\dfrac{1-\sqrt{39}}{2}\approx -2.6218,\qquad \dfrac{1+\sqrt{39}}{2}\approx 3.6218.$$

Because the coefficient of $$p^{2}$$ is positive, inequality (i) holds for

$$p\lt \dfrac{1-\sqrt{39}}{2}\qquad\text{or}\qquad p\gt \dfrac{1+\sqrt{39}}{2}.$$

Next, solve (ii):

$$2p^{2}-2p-119=0 \;\;\Longrightarrow\;\; p=\dfrac{2\pm\sqrt{(-2)^{2}-4\cdot2\cdot(-119)}}{4} =\dfrac{2\pm2\sqrt{239}}{4} =\dfrac{1\pm\sqrt{239}}{2}.$$

Numerically,

$$\dfrac{1-\sqrt{239}}{2}\approx -7.2295,\qquad \dfrac{1+\sqrt{239}}{2}\approx 8.2295.$$

Since here the quadratic opens upward and the inequality is “$$\le 0$$”, condition (ii) is satisfied for

$$\dfrac{1-\sqrt{239}}{2}\le p\le\dfrac{1+\sqrt{239}}{2}.$$

The required $$p$$ must satisfy both (i) and (ii) simultaneously. Intersecting the two solution sets gives two disjoint intervals:

$$\Bigl[\dfrac{1-\sqrt{239}}{2},\,\dfrac{1-\sqrt{39}}{2}\Bigr) \;\;\;\text{and}\;\;\; \Bigl(\dfrac{1+\sqrt{39}}{2},\,\dfrac{1+\sqrt{239}}{2}\Bigr].$$

Numerically these are

$$[-7.2295,\,-2.6218)\quad\text{and}\quad(3.6218,\,8.2295].$$

We now form the set

$$S=\{q:q=p^{2} \text{ and } q \text{ is an integer}\}.$$

For any $$q$$ in $$S$$ either $$p=\sqrt{q}$$ or $$p=-\sqrt{q}$$ must lie in at least one of the above intervals.

For $$p\lt 0$$ we require

$$-\sqrt{q}\in[-7.2295,\,-2.6218) \;\;\Longleftrightarrow\;\; \sqrt{q}\in(2.6218,\,7.2295].$$

This implies

$$6.87\lt q\le 52.25,$$

hence all integers

$$q=7,8,9,\ldots,52.$$

For $$p\gt 0$$ we require

$$\sqrt{q}\in(3.6218,\,8.2295],$$

that is

$$13.12\lt q\le 67.77,$$

whence all integers

$$q=14,15,16,\ldots,67.$$

The union of the two integer ranges is simply the continuous block

$$q=7,8,9,\ldots,67.$$

Counting the elements, we get

$$67-7+1=61.$$

So, the answer is $$61$$.

We have the two circles

$$\bigl(x-1\bigr)^2+\bigl(y-1\bigr)^2 = 1$$

with centre $$C_1(1,\,1)$$ and radius $$r_1 = 1,$$ and

$$\bigl(x-9\bigr)^2+\bigl(y-1\bigr)^2 = 4$$

with centre $$C_2(9,\,1)$$ and radius $$r_2 = 2.$$

The variable straight line is

$$3x+4y=\alpha.$$

Formula used: the perpendicular distance of a point $$(x_0,\,y_0)$$ from the line $$Ax+By+C=0$$ is

$$\dfrac{|Ax_0+By_0+C|}{\sqrt{A^2+B^2}}.$$

Writing the given line in the form $$3x+4y-\alpha=0,$$ its coefficients are $$A=3,$$ $$B=4,$$ $$C=-\alpha,$$ and $$\sqrt{A^2+B^2}=\sqrt{9+16}=5.$$

Perpendicular distance of $$C_1(1,1)$$ from the line is therefore

$$d_1=\dfrac{|3\cdot1+4\cdot1-\alpha|}{5}= \dfrac{|7-\alpha|}{5}.$$

Similarly, distance of $$C_2(9,1)$$ from the line is

$$d_2=\dfrac{|3\cdot9+4\cdot1-\alpha|}{5}= \dfrac{|31-\alpha|}{5}.$$

The line must not cut either circle; it may at the most touch them. Hence we need

$$d_1\ge r_1 \quad\text{and}\quad d_2\ge r_2,$$

that is

$$\dfrac{|7-\alpha|}{5}\;\ge\;1\quad\Longrightarrow\quad |7-\alpha|\;\ge\;5,$$

$$\dfrac{|31-\alpha|}{5}\;\ge\;2\quad\Longrightarrow\quad |31-\alpha|\;\ge\;10.$$

Simplifying the first inequality gives

$$\alpha\le 2 \quad\text{or}\quad \alpha\ge 12,$$

while the second gives

$$\alpha\le 21 \quad\text{or}\quad \alpha\ge 41.$$

Combining them we obtain three possible ranges

$$(-\infty,\,2],\qquad [12,\,21],\qquad [41,\,\infty).$$

Next, for the line to lie between the two circles, the centres $$C_1$$ and $$C_2$$ must lie on opposite sides of the line. This happens exactly when

$$(7-\alpha)(31-\alpha)\lt 0,$$

because the expressions $$7-\alpha$$ and $$31-\alpha$$ are, up to sign, the numerators of the two distances and indicate on which side of the line each centre lies.

The inequality

$$(7-\alpha)(31-\alpha)\lt 0$$

is satisfied whenever

$$7\lt \alpha\lt 31.$$

Intersecting this with the three ranges obtained earlier, only the middle interval survives:

$$12\le\alpha\le21.$$

Thus every real number $$\alpha$$ in the closed interval $$[12,\,21]$$ (including the endpoints, which correspond to tangency) places the line between the two circles without cutting them.

The integral values are

$$12,\,13,\,14,\,15,\,16,\,17,\,18,\,19,\,20,\,21.$$

Count: $$10$$ integers.

Sum: Using the formula for the sum of an arithmetic progression,

$$S=\dfrac{n}{2}(a_1+a_n)=\dfrac{10}{2}(12+21)=5\times33=165.$$

Hence, the correct answer is Option A: $$165.$$

We have the circle

$$x^{2}+y^{2}-2x+4y+1=0.$$

The general form of a circle is written as

$$x^{2}+y^{2}+2gx+2fy+c=0,$$

so by comparing we see

$$2g=-2 \;\;\Longrightarrow\;\; g=-1,$$

$$2f=4 \;\;\Longrightarrow\;\; f=2,$$

$$c=1.$$

Hence the centre of the circle is

$$B(-g,\,-f)=(1,\,-2).$$

The two tangents drawn from the external point

$$A(3,\,1)$$

touch the circle at the points $$P$$ and $$Q$$. The line $$PQ$$ is therefore the chord of contact of $$A$$ with respect to the given circle.

The formula for the chord of contact from an external point $$(x_{1},y_{1})$$ to the circle $$x^{2}+y^{2}+2gx+2fy+c=0$$ is obtained by replacing $$x^{2}$$ by $$xx_{1}$$ and $$y^{2}$$ by $$yy_{1}$$ in the equation of the circle. Symbolically,

$$T=0:\; xx_{1}+yy_{1}+g(x+x_{1})+f(y+y_{1})+c=0.$$

Here $$(x_{1},y_{1})=(3,\,1)$$. Substituting $$g=-1,\;f=2,\;c=1$$ we get

$$xx_{1}+yy_{1}+g(x+x_{1})+f(y+y_{1})+c=0,$$

$$x\cdot3+y\cdot1+(-1)(x+3)+2(y+1)+1=0.$$

Simplifying term by term:

$$3x+y-x-3+2y+2+1=0,$$

$$(3x-x)=2x,\quad (y+2y)=3y,\quad (-3+2+1)=0,$$

so the chord of contact is

$$2x+3y=0.$$

Thus the straight line joining $$P$$ and $$Q$$ is

$$PQ:\;2x+3y=0.$$

Both triangles $$\triangle APQ$$ and $$\triangle BPQ$$ have the same base $$PQ$$. Therefore the ratio of their areas equals the ratio of the perpendicular distances of the vertices $$A$$ and $$B$$ from the line $$PQ$$. That is,

$$\frac{\text{area }\triangle APQ}{\text{area }\triangle BPQ} =\frac{\text{distance of }A\text{ from }PQ} {\text{distance of }B\text{ from }PQ}.$$

The distance of a point $$(x_{0},y_{0})$$ from the line $$2x+3y=0$$ is given by the point-to-line distance formula:

$$d=\frac{|2x_{0}+3y_{0}|}{\sqrt{2^{2}+3^{2}}} =\frac{|2x_{0}+3y_{0}|}{\sqrt{13}}.$$

Distance of $$A(3,1)$$ from $$PQ$$

$$d_{A}=\frac{|2\cdot3+3\cdot1|}{\sqrt{13}} =\frac{|6+3|}{\sqrt{13}} =\frac{9}{\sqrt{13}}.$$

Distance of $$B(1,-2)$$ from $$PQ$$

$$d_{B}=\frac{|2\cdot1+3\cdot(-2)|}{\sqrt{13}} =\frac{|2-6|}{\sqrt{13}} =\frac{4}{\sqrt{13}}.$$

Hence

$$\frac{\text{area }\triangle APQ}{\text{area }\triangle BPQ} =\frac{d_{A}}{d_{B}} =\frac{9/\sqrt{13}}{4/\sqrt{13}} =\frac{9}{4}.$$

We are asked to find

$$8\cdot\frac{\text{area }\triangle APQ}{\text{area }\triangle BPQ} =8\cdot\frac{9}{4} =\frac{8\times9}{4} =2\times9 =18.$$

So, the answer is $$18$$.

Let the moving point be $$P(x,y)$$. We first write the square of its distance from each of the four given points.

Distance-square from $$(0,0)$$ is $$x^{2}+y^{2}$$.

Distance-square from $$(1,0)$$ is $$(x-1)^{2}+y^{2}$$.

Distance-square from $$(0,1)$$ is $$x^{2}+(y-1)^{2}$$.

Distance-square from $$(1,1)$$ is $$(x-1)^{2}+(y-1)^{2}$$.

The question states that the sum of these four expressions equals $$18$$. Hence we have

$$x^{2}+y^{2}+(x-1)^{2}+y^{2}+x^{2}+(y-1)^{2}+(x-1)^{2}+(y-1)^{2}=18.$$

Now we expand every bracket carefully:

$$x^{2}+y^{2}+x^{2}-2x+1+y^{2}+x^{2}+y^{2}-2y+1+x^{2}-2x+1+y^{2}-2y+1=18.$$

Next we collect like terms.

• For $$x^{2}$$: there are four of them, so $$4x^{2}.$$
• For $$y^{2}$$: again four of them, so $$4y^{2}.$$
• For $$x$$: the coefficients are $$-2x-2x=-4x.$$
• For $$y$$: the coefficients are $$-2y-2y=-4y.$$
• Constant terms: $$1+1+1+1=4.$$

Therefore the equation becomes

$$4x^{2}+4y^{2}-4x-4y+4=18.$$

We divide the whole equation by $$4$$ to simplify:

$$x^{2}+y^{2}-x-y+1=\dfrac{18}{4}=4.5.$$

So

$$x^{2}-x+y^{2}-y=4.5-1=3.5.$$

To recognise the locus, we complete the square in both $$x$$ and $$y$$. We recall the algebraic identity $$a^{2}-2ab+b^{2}=(a-b)^{2}$$. For completing the square in an expression of the form $$u^{2}-u$$, we add and subtract $$\left(\dfrac12\right)^{2}=0.25.$$

We add $$0.25$$ for the $$x$$-terms and another $$0.25$$ for the $$y$$-terms on both sides:

$$x^{2}-x+0.25+y^{2}-y+0.25=3.5+0.25+0.25.$$

Simplifying each side gives

$$\bigl(x-0.5\bigr)^{2}+\bigl(y-0.5\bigr)^{2}=4.$$

This is the standard form of the equation of a circle: $$\bigl(x-h\bigr)^{2}+\bigl(y-k\bigr)^{2}=r^{2}$$, where $$(h,k)$$ is the centre and $$r$$ the radius. Comparing, we have centre $$(0.5,0.5)$$ and radius $$r=2$$ because $$r^{2}=4.$

The diameter $$d$$ of the circle is twice the radius:

$$d=2r=2\times2=4.$$

Finally, the square of the diameter is

$$d^{2}=4^{2}=16.$$

So, the answer is $$16$$.

We are told that the two circles have equal radii $$r = 5$$ and touch each other at the point $$P(1,\,2)$$. Their common tangent at the point of contact is given by the straight-line equation $$4x + 3y = 10$$.

For any circle, the radius drawn to the point of tangency is perpendicular to the tangent. Hence the vector along the radius $$\overrightarrow{PC}$$ must be parallel to the normal vector of the tangent. Because the tangent is $$4x + 3y - 10 = 0$$, its normal vector is $$(4,\,3)$$. Therefore every centre lies on the line that passes through $$P(1,\,2)$$ in the direction of $$(4,\,3)$$.

Writing the parametric form of this line, we have $$x = 1 + 4t,\qquad y = 2 + 3t \;,$$ where $$t$$ is a real parameter. Eliminating $$t$$ gives the symmetric form $$\frac{x - 1}{4} = \frac{y - 2}{3}.$$ This is the straight line on which both centres must lie.

Next we use the fact that the distance between a centre and the point of contact is exactly the radius. The distance formula between two points $$(x_1,\,y_1)$$ and $$(x_2,\,y_2)$$ is $$\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}.$$ Setting this distance equal to 5, we obtain $$\sqrt{\,(x - 1)^2 + (y - 2)^2\,} = 5.$$

Because the centres lie on the line found above, let us substitute the parametric coordinates $$x = 1 + 4t,\; y = 2 + 3t$$ into the distance condition:

$$\sqrt{\,(1 + 4t - 1)^2 + (2 + 3t - 2)^2\,} = \sqrt{\,(4t)^2 + (3t)^2\,} = \sqrt{\,16t^2 + 9t^2\,} = \sqrt{25t^2} = 5|t|.$$

We set this equal to 5, obtaining $$5|t| = 5 \;\Longrightarrow\; |t| = 1.$$ Thus the parameter has two possible values: $$t = 1$$ and $$t = -1$$. These correspond to the two different centres situated on opposite sides of the point of contact.

For $$t = 1$$, we find $$C_1(\alpha,\beta) = (1 + 4\cdot1,\; 2 + 3\cdot1) = (5,\,5).$$ Hence $$\alpha = 5,\; \beta = 5.$$

For $$t = -1$$, we find $$C_2(\gamma,\delta) = (1 + 4\cdot(-1),\; 2 + 3\cdot(-1)) = (-3,\,-1).$$ Hence $$\gamma = -3,\; \delta = -1.$$

We are required to evaluate the absolute value $$\left|(\alpha + \beta)(\gamma + \delta)\right|.$$ First compute each sum:

$$\alpha + \beta = 5 + 5 = 10,$$ $$\gamma + \delta = (-3) + (-1) = -4.$$

Multiplying, we get $$(\alpha + \beta)(\gamma + \delta) = 10 \times (-4) = -40.$$ Taking the absolute value, $$\left| -40 \right| = 40.$$

So, the answer is $$40$$.

Let $$P = (x, y)$$. The condition is that the distance from $$P$$ to $$(5, 0)$$ is three times the distance from $$P$$ to $$(-5, 0)$$. Squaring both sides: $$(x - 5)^2 + y^2 = 9[(x + 5)^2 + y^2]$$.

Expanding: $$x^2 - 10x + 25 + y^2 = 9x^2 + 90x + 225 + 9y^2$$. Rearranging: $$8x^2 + 100x + 8y^2 + 200 = 0$$, which simplifies to $$x^2 + \frac{25}{2}x + y^2 + 25 = 0$$.

Completing the square: $$\left(x + \frac{25}{4}\right)^2 + y^2 = \frac{625}{16} - 25 = \frac{625 - 400}{16} = \frac{225}{16}$$.

This is a circle with radius $$r = \frac{15}{4}$$, so $$r^2 = \frac{225}{16}$$ and $$4r^2 = \frac{225}{4} = 56.25$$.

Rounding to the nearest integer, $$4r^2 = 56$$.

The correct answer is $$56$$.

If all normals to a curve pass through a fixed point $$(a, b)$$, then the curve is a circle with center $$(a, b)$$. Let the equation be $$(x - a)^2 + (y - b)^2 = r^2$$.

Since the curve passes through $$(3, -3)$$ and $$(4, -2\sqrt{2})$$, we have:

$$(3 - a)^2 + (-3 - b)^2 = (4 - a)^2 + (-2\sqrt{2} - b)^2$$

Expanding: $$9 - 6a + a^2 + 9 + 6b + b^2 = 16 - 8a + a^2 + 8 + 4\sqrt{2}b + b^2$$.

Simplifying: $$18 - 6a + 6b = 24 - 8a + 4\sqrt{2}b$$, which gives $$2a + (6 - 4\sqrt{2})b = 6$$.

We are also given $$a - 2\sqrt{2}b = 3$$, so $$a = 3 + 2\sqrt{2}b$$. Substituting into the first equation:

$$2(3 + 2\sqrt{2}b) + (6 - 4\sqrt{2})b = 6$$, which gives $$6 + 4\sqrt{2}b + 6b - 4\sqrt{2}b = 6$$, so $$6b = 0$$, hence $$b = 0$$.

Then $$a = 3 + 0 = 3$$, and $$a^2 + b^2 + ab = 9 + 0 + 0 = 9$$.

The circle is $$(x - 2)^2 + (y - 3)^2 = 25$$ with center $$C = (2, 3)$$. The given point is $$(5, 7)$$, which lies on the circle since $$(5-2)^2 + (7-3)^2 = 9 + 16 = 25$$.

The slope of the radius from the center $$(2, 3)$$ to $$(5, 7)$$ is $$\frac{7 - 3}{5 - 2} = \frac{4}{3}$$. The normal to the circle at $$(5, 7)$$ passes through the center and has slope $$\frac{4}{3}$$, so its equation is $$y - 7 = \frac{4}{3}(x - 5)$$. Setting $$y = 0$$: $$-7 = \frac{4}{3}(x - 5)$$, giving $$x = 5 - \frac{21}{4} = -\frac{1}{4}$$. The normal meets the $$x$$-axis at $$\left(-\frac{1}{4}, 0\right)$$.

The tangent at $$(5, 7)$$ is perpendicular to the radius, so it has slope $$-\frac{3}{4}$$. Its equation is $$y - 7 = -\frac{3}{4}(x - 5)$$. Setting $$y = 0$$: $$-7 = -\frac{3}{4}(x - 5)$$, giving $$x = 5 + \frac{28}{3} = \frac{43}{3}$$. The tangent meets the $$x$$-axis at $$\left(\frac{43}{3}, 0\right)$$.

The triangle is formed by the point $$(5, 7)$$, the $$x$$-intercept of the normal $$\left(-\frac{1}{4}, 0\right)$$, and the $$x$$-intercept of the tangent $$\left(\frac{43}{3}, 0\right)$$. The base along the $$x$$-axis has length $$\frac{43}{3} - \left(-\frac{1}{4}\right) = \frac{43}{3} + \frac{1}{4} = \frac{172 + 3}{12} = \frac{175}{12}$$, and the height is $$7$$.

The area is $$A = \frac{1}{2} \times \frac{175}{12} \times 7 = \frac{1225}{24}$$. Therefore, $$24A = 1225$$.

The correct answer is $$1225$$.

We are told that a line $$y = mx + c$$ is tangent to the circle $$(x - 3)^2 + y^2 = 1$$ and that this line is perpendicular to another line $$L_1$$, which itself is the tangent to the circle $$x^2 + y^2 = 1$$ at the point $$\Bigl(\dfrac1{\sqrt2}, \dfrac1{\sqrt2}\Bigr)$$. Our task is to determine the relation satisfied by the constant $$c$$.

We begin with the tangent $$L_1$$ to the circle $$x^2 + y^2 = 1$$. A standard result for a circle of the form $$x^2 + y^2 = r^2$$ is that the tangent at any point $$(x_1,y_1)$$ on the circle is given by $$x x_1 + y y_1 = r^2$$. Here $$r = 1$$ and the given point is $$\Bigl(\dfrac1{\sqrt2}, \dfrac1{\sqrt2}\Bigr)$$, so the tangent is

$$x\Bigl(\dfrac1{\sqrt2}\Bigr) + y\Bigl(\dfrac1{\sqrt2}\Bigr) = 1.$$

Multiplying by $$\sqrt2$$ gives $$x + y = \sqrt2$$, which we rewrite in slope-intercept form:

$$y = -x + \sqrt2.$$

Thus the slope of $$L_1$$ is $$m_1 = -1$$.

The desired tangent $$y = mx + c$$ must be perpendicular to $$L_1$$, and for two lines to be perpendicular, the product of their slopes is $$-1$$. Therefore we have

$$m \cdot (-1) = -1 \;\;\Longrightarrow\;\; m = 1.$$

Hence the required tangent has the form $$y = x + c$$.

Next we impose the condition that this line is tangent to the circle $$(x - 3)^2 + y^2 = 1$$. To apply the tangency condition, we substitute $$y = x + c$$ into the circle’s equation and require that the resulting quadratic in $$x$$ have exactly one real root; equivalently, its discriminant must be zero.

Substituting, we obtain

$$\bigl(x - 3\bigr)^2 + \bigl(x + c\bigr)^2 = 1.$$

Expanding each square gives

$$(x^2 - 6x + 9) + (x^2 + 2cx + c^2) = 1.$$

Combining like terms yields

$$2x^2 + (2c - 6)x + (c^2 + 9) = 1.$$

Bringing the 1 to the left side, we have

$$2x^2 + (2c - 6)x + (c^2 + 9 - 1) = 0,$$

which simplifies to

$$2x^2 + (2c - 6)x + (c^2 + 8) = 0.$$

This is a quadratic equation in $$x$$ of the form $$Ax^2 + Bx + C = 0$$ with coefficients $$A = 2$$, $$B = 2c - 6$$ and $$C = c^2 + 8$$. For tangency we set the discriminant $$\Delta = B^2 - 4AC$$ equal to zero. Thus

$$\bigl(2c - 6\bigr)^2 - 4 \cdot 2 \cdot \bigl(c^2 + 8\bigr) = 0.$$

First, compute $$\bigl(2c - 6\bigr)^2$$:

$$\bigl(2c - 6\bigr)^2 = 4(c - 3)^2 = 4(c^2 - 6c + 9) = 4c^2 - 24c + 36.$$

Next, compute $$4 \cdot 2 \cdot \bigl(c^2 + 8\bigr)$$:

$$4 \cdot 2 \cdot \bigl(c^2 + 8\bigr) = 8c^2 + 64.$$

Setting the discriminant to zero gives

$$4c^2 - 24c + 36 - \bigl(8c^2 + 64\bigr) = 0.$$

Simplifying term by term:

$$4c^2 - 24c + 36 - 8c^2 - 64 = 0$$

$$-4c^2 - 24c - 28 = 0.$$

Multiplying both sides by $$-1$$ to make the leading coefficient positive, we obtain

$$4c^2 + 24c + 28 = 0.$$

Finally, divide every term by $$4$$ to reduce the equation to a monic quadratic:

$$c^2 + 6c + 7 = 0.$$

This matches Option C in the list. Hence, the correct answer is Option C.

We start by noting that the circle touches the y-axis at the point $$\,(0,4)\,.$$ Because the y-axis is the vertical line $$x=0,$$ tangency implies two things. First, the perpendicular from the centre to the y-axis is horizontal, so the centre must have the same y-coordinate as the point of contact; that is, the centre is $$\,(a,4)\,.$$ Second, the distance from the centre to the y-axis equals the radius, so the radius satisfies $$r=\lvert a\rvert.$$

The circle also passes through the point $$\,(2,0).$$ Using the distance formula between two points, the squared distance from the centre $$\,(a,4)$$ to this point must equal the squared radius:

$$\bigl(2-a\bigr)^2+\bigl(0-4\bigr)^2 = r^2.$$

We have already observed that $$r^2=a^2,$$ so substituting:

$$\bigl(2-a\bigr)^2+16=a^2.$$

Expanding the square and collecting like terms:

$$4-4a+a^2+16=a^2.$$

The $$a^2$$ terms cancel immediately, leaving

$$20-4a=0.$$

Solving for $$a$$ gives

$$4a=20 \;\;\Longrightarrow\;\; a=5.$$

Hence the centre is $$\,(5,4)$$ and the radius is $$r=\lvert5\rvert=5.$$ The standard form of the circle is therefore

$$\bigl(x-5\bigr)^2+\bigl(y-4\bigr)^2=25.$$

To determine whether a given line is tangent, we use the distance-from-a-point-to-a-line formula. For a line $$Ax+By+C=0,$$ the perpendicular distance from a point $$\,(x_0,y_0)$$ to the line is

$$d=\dfrac{\lvert Ax_0+By_0+C\rvert}{\sqrt{A^2+B^2}}.$$

The line is tangent to the circle precisely when this distance equals the radius $$r=5.$$ We now test each option using the centre $$\,(5,4).$$

Option A: $$4x-3y+17=0$$

Numerator: $$4(5)-3(4)+17=20-12+17=25.$$
Denominator: $$\sqrt{4^2+(-3)^2}=\sqrt{16+9}=5.$$
Distance: $$d=\dfrac{25}{5}=5=r.$$ So the line is a tangent.

Option B: $$3x-4y-24=0$$

Numerator: $$3(5)-4(4)-24=15-16-24=-25.$$
Denominator: $$\sqrt{3^2+(-4)^2}=\sqrt{9+16}=5.$$
Distance: $$d=\dfrac{\lvert-25\rvert}{5}=5=r.$$ So the line is a tangent.

Option C: $$3x+4y-6=0$$

Numerator: $$3(5)+4(4)-6=15+16-6=25.$$
Denominator: $$\sqrt{3^2+4^2}=\sqrt{9+16}=5.$$
Distance: $$d=\dfrac{25}{5}=5=r.$$ So the line is a tangent.

Option D: $$4x+3y-8=0$$

Numerator: $$4(5)+3(4)-8=20+12-8=24.$$
Denominator: $$\sqrt{4^2+3^2}=\sqrt{16+9}=5.$$
Distance: $$d=\dfrac{24}{5}=4.8\neq5.$$ The distance is not equal to the radius, so this line is not a tangent.

Hence, the correct answer is Option 4.

We have the parabola $$y^{2}=4x$$. For any parabola of the form $$y^{2}=4ax$$, the focus is $$(a,0)$$ and the latus-rectum is the line $$x=a$$ whose length is $$4a$$.

Comparing $$y^{2}=4x$$ with $$y^{2}=4ax$$ gives $$a=1$$. Therefore

$$\text{Equation of latus rectum}=x=a=1,$$

$$\text{Length of latus rectum}=4a=4.$$

The points where this line meets the parabola are obtained by substituting $$x=1$$ in $$y^{2}=4x$$:

$$y^{2}=4(1)=4 \;\;\Longrightarrow\;\; y=\pm2.$$

Hence the latus-rectum is the segment joining $$(1,2)$$ and $$(1,-2)$$, so its length is indeed

$$\sqrt{(1-1)^{2}+(2-(-2))^{2}}=\sqrt{0+4^{2}}=4.$$

This segment is given to be the common chord of two equal circles $$C_{1}$$ and $$C_{2}$$, each having radius

$$R=2\sqrt{5}.$$

Let $$O_{1}$$ and $$O_{2}$$ be the centres of $$C_{1}$$ and $$C_{2}$$ respectively, and let $$d$$ be the (perpendicular) distance from either centre to the chord. For a circle, the length $$2\ell$$ of a chord at distance $$d$$ from the centre satisfies

$$\ell^{2}=R^{2}-d^{2}.$$

Here the full chord length is $$4$$, so the half-length is $$\ell=2$$. Substituting $$\ell=2$$ and $$R=2\sqrt{5}$$ gives

$$2^{2}=R^{2}-d^{2} \;\;\Longrightarrow\;\; 4=(2\sqrt{5})^{2}-d^{2}$$

$$4=20-d^{2}$$

$$d^{2}=20-4=16 \;\;\Longrightarrow\;\; d=4.$$

The latus-rectum is the vertical line $$x=1$$, so the perpendicular distance is measured horizontally. Hence the $$x$$-coordinate of each centre is $$4$$ units away from $$x=1$$:

$$x=1+4=5 \quad\text{or}\quad x=1-4=-3.$$

The chord’s midpoint is $$(1,0)$$, lying on the $$x$$-axis, and the circles are symmetric about this axis, so the centres must have $$y=0$$. Therefore

$$O_{1}=(5,0), \qquad O_{2}=(-3,0).$$

The distance between the centres is then

$$|5-(-3)|=\;8.$$

Hence, the correct answer is Option B.

We have to find the centre $$\,(h,k)\,$$ of a circle which

1. passes through the point $$(0,1)$$, and

2. touches (i.e. is tangent to) the parabola $$y=x^{2}$$ at the point $$(2,4)$$.

Whenever a circle is tangent to a curve at some point, the radius drawn to the point of contact is perpendicular to the tangent of the curve at that point. So our first task is to determine the tangent and then the normal (perpendicular) at $$(2,4)$$ on the parabola.

The parabola is $$y=x^{2}$$. Its derivative gives the slope of the tangent:

$$\frac{dy}{dx}=2x.$$

At $$x=2$$, the slope of the tangent is

$$m_{\text{tan}} = 2\cdot2 = 4.$$

The tangent line at $$(2,4)$$ therefore has slope 4. The slope of the normal (perpendicular) is the negative reciprocal of 4, namely $$-\tfrac14$$. Hence the equation of the normal through $$(2,4)$$ is

$$y-4=-\frac14\,(x-2).$$

Our centre $$(h,k)$$ must lie on this normal, so

$$k-4=-\frac14\,(h-2).$$

Multiplying by 4 to clear the denominator,

$$4(k-4)=-(h-2),$$

$$4k-16=-h+2,$$

$$h+4k-18=0.$$ We shall label this as equation $$(1).$$

Next we use the fact that the circle passes through two known points: the point of tangency $$(2,4)$$ and the point $$(0,1)$$. If $$r$$ is the radius, then

$$r^{2}=(2-h)^{2}+(4-k)^{2}\quad\text{and}\quad r^{2}=(h-0)^{2}+(k-1)^{2}.$$

Since both equal $$r^{2}$$, they are equal to each other:

$$(2-h)^{2}+(4-k)^{2}=h^{2}+(k-1)^{2}.$$

Now we expand each square fully, step by step:

$$(2-h)^{2}=h^{2}-4h+4,$$

$$(4-k)^{2}=k^{2}-8k+16,$$

$$(k-1)^{2}=k^{2}-2k+1.$$

Substituting these expansions into the equality gives

$$\bigl(h^{2}-4h+4\bigr)+\bigl(k^{2}-8k+16\bigr)=h^{2}+\bigl(k^{2}-2k+1\bigr).$$

Combining like terms on the left-hand side:

$$h^{2}+k^{2}-4h-8k+20 = h^{2}+k^{2}-2k+1.$$

Now we cancel $$h^{2}+k^{2}$$ from both sides:

$$-4h-8k+20 = -2k+1.$$

Bringing all terms to one side:

$$-4h-8k+20+2k-1=0,$$

$$-4h-6k+19=0.$$

Multiplying by $$-1$$ to make coefficients positive:

$$4h+6k-19=0.$$

This is our second linear relation between $$h$$ and $$k$$; call it equation $$(2).$$

We now solve the simultaneous linear equations:

$$(1)\;:\;h+4k-18=0,$$

$$(2)\;:\;4h+6k-19=0.$$

From equation $$(1)$$ we isolate $$h$$:

$$h=18-4k.$$

Substituting this value of $$h$$ into equation $$(2)$$:

$$4(18-4k)+6k-19=0,$$

$$72-16k+6k-19=0,$$

$$72-10k-19=0,$$

$$53-10k=0,$$

$$10k=53,$$

$$k=\frac{53}{10}.$$

Now we substitute $$k=\frac{53}{10}$$ back into $$h=18-4k$$:

$$h = 18 - 4\left(\frac{53}{10}\right)= 18 - \frac{212}{10} = \frac{180}{10} - \frac{212}{10} = -\frac{32}{10}= -\frac{16}{5}.$$

Thus the centre is

$$\left(h,k\right)=\left(-\frac{16}{5},\;\frac{53}{10}\right).$$

Comparing with the given options, this corresponds to Option D.

Hence, the correct answer is Option D.

The two given circles are

$$x^{2}+y^{2}-6x = 0 \qquad\text{and}\qquad x^{2}+y^{2}-4y = 0.$$

The standard method to obtain every circle that passes through the common points (their “radical family”) is to write

$$\bigl(x^{2}+y^{2}-6x\bigr) \;+\;\lambda\,\bigl(x^{2}+y^{2}-4y\bigr)=0,$$

where $$\lambda$$ is a real parameter. Expanding the brackets we get

$$\;(1+\lambda)\,(x^{2}+y^{2})\;-\;6x\;-\;4\lambda\,y\;=\;0.$$

To identify the centre easily, we divide every term by the common factor $$1+\lambda$$ (allowed as long as $$\lambda\neq-1$$, which will turn out to be true):

$$x^{2}+y^{2}\;-\;\dfrac{6}{1+\lambda}\,x\;-\;\dfrac{4\lambda}{1+\lambda}\,y = 0.$$

A circle in general form is written as

$$x^{2}+y^{2}-2gx-2fy+c=0,$$

whose centre is $$\bigl(g,f\bigr).$$ Comparing coefficients, we obtain

$$2g=\dfrac{6}{1+\lambda}\;\Longrightarrow\;g=\dfrac{3}{1+\lambda},$$

$$2f=\dfrac{4\lambda}{1+\lambda}\;\Longrightarrow\;f=\dfrac{2\lambda}{1+\lambda}.$$

Therefore the centre of the required circle is

$$\left(\dfrac{3}{1+\lambda},\;\dfrac{2\lambda}{1+\lambda}\right).$$

It is given that this centre lies on the straight line

$$2x-3y+12=0.$$

Substituting $$x=\dfrac{3}{1+\lambda}$$ and $$y=\dfrac{2\lambda}{1+\lambda}$$ into the line’s equation, we have

$$2\left(\dfrac{3}{1+\lambda}\right)-3\left(\dfrac{2\lambda}{1+\lambda}\right)+12=0.$$

The denominators are the same, so we combine the fractions:

$$\dfrac{6-6\lambda}{1+\lambda}+12=0.$$

To clear the denominator, multiply every term by $$1+\lambda$$:

$$6-6\lambda+12(1+\lambda)=0.$$

Expanding the last bracket,

$$6-6\lambda+12+12\lambda = 0.$$

Collecting like terms,

$$(6+12)+( -6\lambda+12\lambda)=0 \;\;\Longrightarrow\;\; 18+6\lambda=0.$$

Dividing by $$6$$ we find

$$\lambda=-3.$$

Now we substitute $$\lambda=-3$$ back into the family to get the explicit equation of the desired circle. First compute $$1+\lambda=1-3=-2.$$ Then

$$(1+\lambda)(x^{2}+y^{2})-6x-4\lambda y=0$$ $$\Longrightarrow\;(-2)(x^{2}+y^{2})-6x-4(-3)y=0$$ $$\Longrightarrow\;-2x^{2}-2y^{2}-6x+12y=0.$$

Multiplying by $$-1$$ to simplify,

$$2x^{2}+2y^{2}+6x-12y=0,$$

and dividing by $$2$$ gives

$$x^{2}+y^{2}+3x-6y=0.$$

Any point lying on this circle must satisfy the equation above. We now test the four options.

Option A: $$(-1,3)$$ $$(-1)^{2}+3^{2}+3(-1)-6(3)=1+9-3-18 = -11\neq0.$$

Option B: $$(-3,6)$$ $$(-3)^{2}+6^{2}+3(-3)-6(6)=9+36-9-36 = 0.$$

Option C: $$(-3,1)$$ $$(-3)^{2}+1^{2}+3(-3)-6(1)=9+1-9-6 = -5\neq0.$$

Option D: $$(1,-3)$$ $$1^{2}+(-3)^{2}+3(1)-6(-3)=1+9+3+18 = 31\neq0.$$

Only the point $$(-3,6)$$ satisfies the circle’s equation.

Hence, the correct answer is Option B.

We are given the circle $$x^{2}+y^{2}=r^{2}$$, whose centre is clearly at the origin $$(0,0)$$ and whose radius is $$r$$.

The chord lies on the straight line $$y-2x=3$$. Re-write this line in the standard linear form $$Ax+By+C=0$$, because the perpendicular-distance formula uses that form:

$$y-2x-3=0\;\;\Longrightarrow\;\;-2x+y-3=0.$$

For convenience we multiply by $$-1$$ so that the coefficient of $$x$$ is positive:

$$2x-y+3=0.$$

The perpendicular distance $$d$$ from a point $$(x_{0},y_{0})$$ to the line $$Ax+By+C=0$$ is given by the formula

$$d=\frac{|Ax_{0}+By_{0}+C|}{\sqrt{A^{2}+B^{2}}}.$$

Here our point is the centre $$(0,0)$$, and the coefficients are $$A=2,\;B=-1,\;C=3$$. Substituting, we obtain

$$d=\frac{|\,2\cdot0+(-1)\cdot0+3\,|}{\sqrt{2^{2}+(-1)^{2}}} =\frac{|\,3\,|}{\sqrt{4+1}} =\frac{3}{\sqrt{5}}.$$

So,

$$d^{2}=\left(\frac{3}{\sqrt{5}}\right)^{2}=\frac{9}{5}.$$

Now recall the chord-length formula in a circle: if a chord is at a perpendicular distance $$d$$ from the centre and the radius of the circle is $$r$$, then its length $$L$$ is

$$L=2\sqrt{\,r^{2}-d^{2}\,}.$$

According to the problem, the length of this chord is exactly $$r$$ itself. Hence we set

$$r=2\sqrt{\,r^{2}-d^{2}\,}.$$

Square both sides to eliminate the square root:

$$r^{2}=4\bigl(r^{2}-d^{2}\bigr).$$

Expand the right side and gather like terms:

$$r^{2}=4r^{2}-4d^{2}\;\;\Longrightarrow\;\;4d^{2}=4r^{2}-r^{2}=3r^{2}.$$

Divide by $$3$$ to isolate $$r^{2}$$:

$$r^{2}=\frac{4}{3}d^{2}.$$

We already found $$d^{2}=\dfrac{9}{5}$$, so substitute:

$$r^{2}=\frac{4}{3}\times\frac{9}{5} =\frac{36}{15} =\frac{12}{5}.$$

Hence, the correct answer is Option D.

We have the circle $$x^{2}+y^{2}-8x-4y+16=0$$. To recognise its centre and radius, we complete squares.

Re-grouping the $$x$$ and $$y$$ terms,

$$x^{2}-8x+y^{2}-4y+16=0.$$

Inside each bracket we add and subtract the square of half the coefficient:

$$\bigl(x^{2}-8x+16\bigr)+\bigl(y^{2}-4y+4\bigr)+16-16-4=0.$$

This simplifies to

$$\left(x-4\right)^{2}+\left(y-2\right)^{2}-4=0,$$

so in standard form

$$\left(x-4\right)^{2}+\left(y-2\right)^{2}=4.$$

Hence the centre is $$C(4,\,2)$$ and the radius is $$r=2$$.

Let $$O(0,\,0)$$ be the origin. We first find the distance $$OC$$:

$$OC=\sqrt{4^{2}+2^{2}}=\sqrt{16+4}=2\sqrt5.$$

For any external point, the square of the length of the tangent drawn to a circle equals the power of the point, namely

$$\text{(tangent length)}^{2}=OC^{2}-r^{2}.$$

Substituting $$OC^{2}=20$$ and $$r^{2}=4$$ gives

$$OA^{2}=OB^{2}=20-4=16\;\Longrightarrow\;OA=OB=4.$$

The tangents touch the circle at $$A$$ and $$B$$, and the radii $$CA$$ and $$CB$$ are perpendicular to the tangents. Thus each of the triangles $$\triangle OCA$$ and $$\triangle OCB$$ is right-angled at $$A$$ and $$B$$ respectively.

Let $$\alpha=\angle OCA$$. In $$\triangle OCA$$ we know three sides: $$OC=2\sqrt5,\; CA=r=2,\; OA=4.$$ Applying the Cosine Rule (stated: in any triangle $$a^{2}=b^{2}+c^{2}-2bc\cos A$$), with side $$OA$$ opposite angle $$\alpha$$, we get

$$OA^{2}=OC^{2}+CA^{2}-2\,(OC)\,(CA)\cos\alpha.$$

Substituting the lengths,

$$16=20+4-2\,(2\sqrt5)\,(2)\cos\alpha.$$

Simplifying,

$$16=24-8\sqrt5\,\cos\alpha,$$

$$8\sqrt5\,\cos\alpha=24-16=8,$$

$$\cos\alpha=\dfrac{8}{8\sqrt5}=\dfrac1{\sqrt5}.$$

Therefore

$$\sin\alpha=\sqrt{1-\cos^{2}\alpha}=\sqrt{1-\dfrac1{5}}=\sqrt{\dfrac45}=\dfrac{2}{\sqrt5}.$$

The angle between the two radii $$CA$$ and $$CB$$ is $$\angle ACB=2\alpha$$ because the triangles $$\triangle OCA$$ and $$\triangle OCB$$ are mirror images about $$OC$$. For a circle, the length of a chord subtending an angle $$\theta$$ at the centre is

$$\text{chord}=2r\sin\frac{\theta}{2}.$$

Here the chord is $$AB$$, the subtended angle is $$\theta=2\alpha$$, so

$$AB = 2r\sin\frac{2\alpha}{2}=2r\sin\alpha.$$

Substituting $$r=2$$ and $$\sin\alpha=\dfrac{2}{\sqrt5}$$,

$$AB = 2\,(2)\,\dfrac{2}{\sqrt5}= \dfrac{8}{\sqrt5}.$$

Finally, the required quantity is

$$\bigl(AB\bigr)^{2}= \left(\dfrac{8}{\sqrt5}\right)^{2}= \dfrac{64}{5}.$$

Hence, the correct answer is Option C.

We have the circle $$x^2 + y^2 = 9$$ whose centre is clearly the origin $$O(0,0)$$ and whose radius is $$3$$ because in the standard form $$x^2 + y^2 = r^2$$ the constant term equals $$r^2$$.

Let the end-points of a diameter be $$P(x_1 , y_1)$$ and $$Q(x_2 , y_2)$$. By the very definition of a diameter, the centre is the mid-point of $$PQ$$, so $$\dfrac{x_1 + x_2}{2}=0$$ and $$\dfrac{y_1 + y_2}{2}=0$$. Hence $$x_2 = -x_1$$ and $$y_2 = -y_1$$. Thus we may write

$$P(x_1 , y_1), \qquad Q(-x_1 , -y_1).$$

Because $$P$$ lies on the circle, it satisfies the equation of the circle:

$$x_1^2 + y_1^2 = 9.$$

The given straight line is $$x + y = 2$$. For distance calculations we write it in the standard form $$Ax + By + C = 0$$:

$$x + y - 2 = 0,$$

so $$A = 1, \; B = 1, \; C = -2$$.

Now we recall the perpendicular distance formula.
Formula: The distance from a point $$(x_0 , y_0)$$ to the line $$Ax + By + C = 0$$ is $$ \text{Distance} \;=\; \dfrac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}}. $$

Using this formula, the length of the perpendicular from $$P(x_1 , y_1)$$ to the line $$x + y - 2 = 0$$ is

$$ \alpha \;=\; \dfrac{|1\cdot x_1 + 1\cdot y_1 - 2|}{\sqrt{1^2 + 1^2}} \;=\; \dfrac{|x_1 + y_1 - 2|}{\sqrt{2}}. $$

Similarly, the perpendicular distance from $$Q(-x_1 , -y_1)$$ to the same line is

$$ \beta \;=\; \dfrac{|1(-x_1) + 1(-y_1) - 2|}{\sqrt{2}} \;=\; \dfrac{|\, -x_1 - y_1 - 2|}{\sqrt{2}} \;=\; \dfrac{|x_1 + y_1 + 2|}{\sqrt{2}}. $$

We multiply the two distances to get

$$ \alpha\beta \;=\; \left(\dfrac{|x_1 + y_1 - 2|}{\sqrt{2}}\right) \left(\dfrac{|x_1 + y_1 + 2|}{\sqrt{2}}\right) \;=\; \dfrac{|x_1 + y_1 - 2| \; |x_1 + y_1 + 2|}{2}. $$

Notice that $$|a-b|\,|a+b| = |a^2 - b^2|$$ for any real numbers. Putting $$a = x_1 + y_1$$ and $$b = 2$$, we get

$$ |x_1 + y_1 - 2|\;|x_1 + y_1 + 2| \;=\; |(x_1 + y_1)^2 - 2^2| \;=\; |(x_1 + y_1)^2 - 4|. $$

Hence

$$ \alpha\beta \;=\; \dfrac{\bigl|(x_1 + y_1)^2 - 4\bigr|}{2}. $$

For convenience let us set

$$ s \;=\; x_1 + y_1. $$

Then the expression to be maximised is

$$ \alpha\beta \;=\; \dfrac{|\,s^2 - 4\,|}{2}. $$

We must now find the range of the quantity $$s = x_1 + y_1$$ subject to the circular condition $$x_1^2 + y_1^2 = 9$$. For any real numbers $$x_1, y_1$$ we have the Cauchy-Schwarz (or simply the AM-QM) inequality

$$ (x_1 + y_1)^2 \;\le\; 2\,(x_1^2 + y_1^2). $$

Substituting the fixed value $$x_1^2 + y_1^2 = 9$$ gives

$$ s^2 \;\le\; 2 \times 9 = 18, \qquad\text{so}\qquad -3\sqrt{2} \;\le\; s \;\le\; 3\sqrt{2}. $$

Thus $$s^2$$ ranges from $$0$$ up to $$18$$.

Now observe that $$\alpha\beta$$ depends only on $$s^2$$, not on the sign of $$s$$, and that

$$ \alpha\beta \;=\; \frac{|s^2 - 4|}{2} \;=\; \begin{cases} \dfrac{4 - s^2}{2}, & 0 \le s^2 \le 4, \\ \dfrac{s^2 - 4}{2}, & 4 \le s^2 \le 18. \end{cases} $$

• For $$0 \le s^2 \le 4$$, the numerator $$4 - s^2$$ decreases as $$s^2$$ increases, so the largest value in this interval occurs at the left end $$s^2 = 0$$, giving $$\dfrac{4 - 0}{2} = 2.$$
• For $$4 \le s^2 \le 18$$, the numerator $$s^2 - 4$$ increases as $$s^2$$ increases, so the largest value in this interval occurs at the right end $$s^2 = 18$$, giving $$\dfrac{18 - 4}{2} = \dfrac{14}{2} = 7.$$

Comparing the two candidates $$2$$ and $$7$$, the maximum of $$\alpha\beta$$ is clearly $$7$$.

Hence, the correct answer is Option 7.

We have the circle

$$x^{2}+y^{2}-2x-4y+4=0$$

First, we put this equation into centre-radius form. We add and subtract the appropriate constants:

$$x^{2}-2x+1+y^{2}-4y+4=0+1+4$$

$$\bigl(x^{2}-2x+1\bigr)+\bigl(y^{2}-4y+4\bigr)=1$$

$$(x-1)^{2}+(y-2)^{2}=1$$

Thus the centre of the circle is $$C(1,\,2)$$ and the radius is $$r=1$$.

The straight line is

$$3x+4y=k.$$

For the line to cut the circle at two distinct points, the perpendicular distance from the centre of the circle to the line must be strictly less than the radius. We now state the distance formula:

For a line $$ax+by+c=0$$ and a point $$(x_{0},y_{0})$$, the perpendicular distance is

$$\displaystyle d=\frac{|ax_{0}+by_{0}+c|}{\sqrt{a^{2}+b^{2}}}\,.$$

First we rewrite the given line in the required form:

$$3x+4y-k=0,$$

so $$a=3,\;b=4,\;c=-k.$$ Substituting the centre $$(1,2)$$ into the formula gives

$$d=\frac{|3(1)+4(2)-k|}{\sqrt{3^{2}+4^{2}}} =\frac{|3+8-k|}{\sqrt{9+16}} =\frac{|11-k|}{5}.$$

For two distinct intersection points we need

$$d<r \;\; \Longrightarrow \;\; \frac{|11-k|}{5}<1.$$

Multiplying both sides by $$5$$ we get

$$|11-k|<5.$$

Now we convert the modulus inequality into a double inequality:

$$-5<11-k<5.$$

Subtracting $$11$$ everywhere,

$$-16<-k<-6.$$

Multiplying through by $$-1$$ (and reversing the inequality signs) gives

$$6<k<16.$$

The integral values satisfying this are

$$k=7,\,8,\,9,\,10,\,11,\,12,\,13,\,14,\,15.$$

Counting them, we find $$9$$ such integers.

So, the answer is $$9$$.

We have two curves

$$x^{2}-6x+y^{2}+8=0$$

and

$$x^{2}-8y+y^{2}+16-k=0,\qquad k\gt 0.$$

Each of these equations represents a circle. To see their centres and radii we first write each circle in standard form by completing the square.

For the first curve, isolate the terms in $$x$$:

$$x^{2}-6x = (x-3)^{2}-9.$$

Substituting this into the first equation gives

$$\bigl[(x-3)^{2}-9\bigr] + y^{2}+8 = 0.$$

Simplifying,

$$(x-3)^{2}+y^{2} -1 = 0,$$

or equivalently,

$$(x-3)^{2}+y^{2}=1.$$

Thus the first circle has centre

$$C_{1} = (3,\,0)$$

and radius

$$r_{1} = 1.$$

Now we treat the second curve. Complete the square in $$y$$:

$$y^{2}-8y = (y-4)^{2}-16.$$

Substituting back,

$$x^{2} + \bigl[(y-4)^{2}-16\bigr] + 16 - k = 0.$$

The constants $$-16$$ and $$+16$$ cancel, so we obtain

$$x^{2} + (y-4)^{2} - k = 0,$$

or

$$x^{2} + (y-4)^{2} = k.$$

Hence the second circle has centre

$$C_{2} = (0,\,4)$$

and radius

$$r_{2} = \sqrt{k}.$$

When two circles touch each other, the distance $$d$$ between their centres satisfies one of the following standard conditions:

• External tangency: $$d = r_{1} + r_{2}.$$

• Internal tangency: $$d = |\,r_{1} - r_{2}\,|.$$

First we compute the centre-centre distance. Using the distance formula,

$$d = \sqrt{(3-0)^{2} + (0-4)^{2}} = \sqrt{9 + 16} = 5.$$

Now we analyse the two possible tangencies.

External tangency. Here

$$r_{1} + r_{2} = d \;\Longrightarrow\; 1 + \sqrt{k} = 5.$$

So

$$\sqrt{k} = 4 \;\Longrightarrow\; k = 16.$$

Internal tangency. Here

$$|\,r_{1} - r_{2}\,| = d.$$

Because $$r_{2} \gt r_{1}$$ (otherwise the left side would be negative), we write

$$r_{2} - r_{1} = 5 \;\Longrightarrow\; \sqrt{k} - 1 = 5.$$

Therefore

$$\sqrt{k} = 6 \;\Longrightarrow\; k = 36.$$

Between the two admissible values $$k = 16$$ and $$k = 36$$, the question asks for the largest value. Clearly,

$$k_{\max} = 36.$$

Hence, the correct answer is Option D.

Let the centre of the required circle be the point $$(h,\,k)$$. Because the centre lies on the straight line $$x + y = 2$$ we can write

$$h + k = 2 \quad\quad (1)$$

We are told that the circle touches the vertical line $$x = 3$$. For any point $$(h,\,k)$$ the perpendicular distance to the line $$x = 3$$ equals the absolute difference of their $$x$$-coordinates, so

$$\text{Distance from }(h,k)\text{ to }x = 3 = |h - 3|.$$

This distance is exactly the radius $$r$$ of the circle, hence

$$r = |h - 3|.$$

Similarly, the circle also touches the horizontal line $$y = 2$$. The perpendicular distance from $$(h,\,k)$$ to this line is the absolute difference of their $$y$$-coordinates, giving

$$\text{Distance from }(h,k)\text{ to }y = 2 = |k - 2|.$$

This distance must be the same radius $$r$$, so we also have

$$r = |k - 2|.$$

Because the centre lies in the first quadrant and we already know from (1) that $$h + k = 2,$$ each of $$h$$ and $$k$$ is positive and less than $$2.$$ That means $$h \lt 2$$ and $$k \lt 2,$$ placing the centre to the left of $$x = 3$$ and below $$y = 2.$$ Therefore the absolute value signs can be removed by inserting negative signs inside, yielding

$$r = 3 - h \quad\text{and}\quad r = 2 - k.$$

Equating these two expressions for the same radius gives

$$3 - h = 2 - k.$$

Simplifying, we obtain

$$-h + k = -1$$

or, more neatly,

$$k - h = -1 \quad\quad (2)$$

Now we have the pair of linear equations

$$\begin{aligned} h + k &= 2 \quad\quad&(1)\\ k - h &= -1 \quad\quad&(2) \end{aligned}$$

Adding equations (1) and (2) eliminates $$h$$:

$$\bigl(h + k\bigr) + \bigl(k - h\bigr) = 2 + (-1) \implies 2k = 1 \implies k = \dfrac12.$$

Substituting $$k = \dfrac12$$ back into equation (1) gives

$$h + \dfrac12 = 2 \implies h = 2 - \dfrac12 = \dfrac32.$$

Thus the centre is $$(h,\,k) = \left(\dfrac32,\,\dfrac12\right).$$

Using either expression for $$r,$$ say $$r = 3 - h,$$ we find

$$r = 3 - \dfrac32 = \dfrac32.$$

The diameter $$D$$ of the circle is twice the radius:

$$D = 2r = 2\left(\dfrac32\right) = 3.$$

Hence, the correct answer is Option 3.

Let us denote the centre of the required circle by the coordinates $$(h,k)$$ and its radius by $$r$$.

Hence the general equation of the circle is

$$ (x-h)^2 + (y-k)^2 = r^2. $$

We translate both given geometric conditions into algebraic equations.

First condition: the circle cuts a chord of length $$4a$$ on the $$x$$-axis. On the $$x$$-axis we have $$y=0$$, so substituting $$y=0$$ in the equation of the circle gives

$$ (x-h)^2 + k^2 = r^2. $$

This is a quadratic in $$x$$ whose roots give the $$x$$-coordinates of the points of intersection with the $$x$$-axis. Solving for $$x$$ gives

$$ x = h \pm \sqrt{\,r^2 - k^2\,}. $$

The distance between these two intersection points is therefore

$$ |(h+\sqrt{r^2-k^2})-(h-\sqrt{r^2-k^2})| = 2\sqrt{\,r^2-k^2\,}. $$

But we are told that this distance equals the given chord length $$4a$$, so

$$ 2\sqrt{\,r^2-k^2\,} = 4a \quad\Longrightarrow\quad \sqrt{\,r^2-k^2\,} = 2a \quad\Longrightarrow\quad r^2 - k^2 = 4a^2. \quad -(1) $$

Second condition: the circle passes through the point on the $$y$$-axis that is at a distance $$2b$$ from the origin. That point has coordinates $$(0,\,2b)$$. Substituting $$(x,y)=(0,2b)$$ in the circle’s equation gives

$$ (0-h)^2 + (2b-k)^2 = r^2 \quad\Longrightarrow\quad h^2 + (2b-k)^2 = r^2. \quad -(2) $$

Now we eliminate $$r^2$$ between equations (1) and (2). From (1) we have $$r^2 = k^2 + 4a^2$$. Putting this value into (2) yields

$$ h^2 + (2b-k)^2 = k^2 + 4a^2. $$

Expanding the square and simplifying step by step:

$$ h^2 + (2b-k)^2 = h^2 + (4b^2 - 4bk + k^2). $$

So the left-hand side becomes

$$ h^2 + 4b^2 - 4bk + k^2. $$

Equating this to the right-hand side $$k^2 + 4a^2$$ and cancelling the identical $$k^2$$ terms on both sides, we obtain

$$ h^2 + 4b^2 - 4bk = 4a^2. $$

Collecting all terms to one side we get the Cartesian relation connecting $$h$$ and $$k$$:

$$ h^2 - 4bk + 4b^2 - 4a^2 = 0. $$

To write the locus, we now replace the particular centre coordinates $$(h,k)$$ by the general point $$(x,y)$$:

$$ x^2 - 4by + 4b^2 - 4a^2 = 0. $$

Rearranging to the more familiar form,

$$ x^2 = 4by - 4(b^2 - a^2). $$

This equation contains the squared term $$x^2$$ but no $$y^2$$ term, which is the defining algebraic characteristic of a parabola (a conic with eccentricity one). Hence the centre of the circle moves along a parabolic curve.

Hence, the correct answer is Option D.

We start with the given circle

$$x^{2}+y^{2}+10x+12y+c=0.$$

To recognise its centre and radius, we complete the squares. Write the quadratic terms together:

$$\bigl(x^{2}+10x\bigr)+\bigl(y^{2}+12y\bigr)= -\,c.$$

Now add and subtract the necessary constants inside each bracket. The coefficient of $$x$$ is $$10$$, so half of it is $$5$$ and its square is $$25$$. For $$y$$, the coefficient is $$12$$, half is $$6$$ and its square is $$36$$. Inserting these we get

$$\bigl(x^{2}+10x+25-25\bigr)+\bigl(y^{2}+12y+36-36\bigr)= -\,c.$$

Group the perfect squares and the constants separately:

$$(x+5)^{2}+(y+6)^{2}-25-36= -\,c.$$

Combine the constants $$-25-36=-61$$ and move them to the right side:

$$(x+5)^{2}+(y+6)^{2}=61-c.$$

Hence the circle is centred at $$(-5,-6)$$ and its radius is

$$R=\sqrt{61-c}.$$

Next, an equilateral triangle is inscribed in this circle. For an equilateral triangle the standard relation between the side length $$a$$ and the circum-radius $$R$$ is

$$R=\frac{a}{\sqrt{3}}$$ (circumradius formula for an equilateral triangle).

The area formula for an equilateral triangle is

$$\text{Area}=\frac{\sqrt{3}}{4}\,a^{2}.$$

We are told that the area equals $$27\sqrt{3}$$, so substituting this value gives

$$\frac{\sqrt{3}}{4}\,a^{2}=27\sqrt{3}.$$

Divide both sides by $$\sqrt{3}$$:

$$\frac{1}{4}\,a^{2}=27.$$

Multiply by $$4$$:

$$a^{2}=108.$$

Take the square root (the side length is positive):

$$a=\sqrt{108}=6\sqrt{3}.$$

Now substitute this $$a$$ into the circumradius formula:

$$R=\frac{a}{\sqrt{3}}=\frac{6\sqrt{3}}{\sqrt{3}}=6.$$

Thus the radius of the given circle must be

$$R=6.$$

But from the circle’s equation we already have

$$R=\sqrt{61-c}.$$

Equate the two expressions for $$R$$:

$$\sqrt{61-c}=6.$$

Square both sides to remove the square root:

$$61-c=36.$$

Solve for $$c$$ by moving terms:

$$c=61-36=25.$$

Hence, the correct answer is Option A.

We have a circle which is tangent to the line $$x = y$$ at the point $$(1,\,1)$$ and which also passes through the point $$(1,\,-3)$$. Our aim is to determine the radius of this circle.

Because a radius drawn to the point of contact is always perpendicular to the tangent line, the centre of the circle must lie on the straight line that is perpendicular to $$x = y$$ at the point $$(1,\,1)$$.

The given tangent $$x = y$$ has slope $$1$$. A line perpendicular to it therefore has slope $$-1$$. Writing the equation of the line with slope $$-1$$ through $$(1,\,1)$$, we start from the point-slope form $$y - y_1 = m\,(x - x_1)$$. Substituting $$m=-1$$ and $$(x_1,\,y_1)=(1,\,1)$$, we get

$$y - 1 = -1\,(x - 1).$$

Simplifying,

$$y - 1 = -x + 1,$$

$$x + y = 2.$$

Hence the centre $$C(h,\,k)$$ of the circle must satisfy the linear relation

$$h + k = 2. \quad -(1)$$

Next, we use the distance formula. The radius $$r$$ is the distance from the centre to any point on the circle. First, taking the point of contact $$(1,\,1)$$, we have

$$r^2 = (h - 1)^2 + (k - 1)^2. \quad -(2)$$

The circle also passes through the point $$(1,\,-3)$$, so the distance from the centre to this point must equal the same radius. Therefore,

$$r^2 = (h - 1)^2 + (k + 3)^2. \quad -(3)$$

Because both right-hand sides equal $$r^2$$, we can equate them directly. Substituting from (2) and (3),

$$(h - 1)^2 + (k + 3)^2 = (h - 1)^2 + (k - 1)^2.$$

The term $$(h - 1)^2$$ appears on both sides, so it cancels, leaving

$$(k + 3)^2 = (k - 1)^2.$$

Now we expand both squares step by step:

$$k^2 + 6k + 9 = k^2 - 2k + 1.$$

Subtracting $$k^2$$ from each side eliminates the quadratic terms:

$$6k + 9 = -2k + 1.$$

Adding $$2k$$ to both sides gathers the $$k$$ terms together:

$$8k + 9 = 1.$$

Subtracting $$9$$ from both sides,

$$8k = -8.$$

Dividing by $$8$$, we find

$$k = -1.$$

Using relation (1), $$h + k = 2$$, we substitute $$k = -1$$ to get

$$h + (-1) = 2,$$

$$h = 3.$$

Thus the centre of the circle is $$C(3,\,-1).$$

Finally, we compute the radius using the distance from the centre to $$(1,\,1)$$. By the distance formula,

$$r = \sqrt{(3 - 1)^2 + (-1 - 1)^2}.$$

Calculating the squares explicitly,

$$r = \sqrt{(2)^2 + (-2)^2} = \sqrt{4 + 4} = \sqrt{8}.$$

Since $$\sqrt{8} = 2\sqrt{2},$$ we have

$$r = 2\sqrt{2}.$$

Hence, the correct answer is Option D.

We have the circle $$x^{2}+y^{2}=4$$ whose centre is the origin $$(0,0)$$ and whose radius is $$2$$ because $$\sqrt{4}=2$$.

The given point on the circle is $$P(\sqrt{3},\,1)$$. Clearly $$\left(\sqrt{3}\right)^{2}+1^{2}=3+1=4$$, so the point lies on the circle.

First we write the equation of the tangent to the circle at this point. For a circle $$x^{2}+y^{2}=r^{2}$$, the tangent at $$\bigl(x_{1},y_{1}\bigr)$$ is given by the standard formula

$$x\,x_{1}+y\,y_{1}=r^{2}.$$

Here $$x_{1}=\sqrt{3},\;y_{1}=1,\;r^{2}=4$$, so we substitute:

$$x\bigl(\sqrt{3}\bigr)+y(1)=4.$$

This simplifies to

$$\sqrt{3}\,x+y=4,$$

or, solving for $$y$$ so that the slope is clear,

$$y=4-\sqrt{3}\,x.$$

Next we find the equation of the normal at the same point. The normal is the line through the centre and the point of contact, so it is simply the radius $$OP$$. The slope of $$OP$$ is

$$m_{OP}=\frac{1-0}{\sqrt{3}-0}=\frac{1}{\sqrt{3}}.$$

Using the point-slope form $$y-y_{1}=m(x-x_{1})$$ with point $$P(\sqrt{3},1)$$ and the above slope, we write

$$y-1=\frac{1}{\sqrt{3}}\bigl(x-\sqrt{3}\bigr).$$

Expanding,

$$y-1=\frac{x}{\sqrt{3}}-\frac{\sqrt{3}}{\sqrt{3}}=\frac{x}{\sqrt{3}}-1.$$

Adding $$1$$ to both sides gives

$$y=\frac{x}{\sqrt{3}}.$$

Thus the normal line is

$$y=\frac{1}{\sqrt{3}}\,x.$$

Now we locate the intercepts of the tangent and the normal with the x-axis \;(set $$y=0$$).

For the tangent $$y=4-\sqrt{3}\,x$$ we put $$y=0$$:

$$0=4-\sqrt{3}\,x\quad\Longrightarrow\quad\sqrt{3}\,x=4\quad\Longrightarrow\quad x=\frac{4}{\sqrt{3}}.$$

Hence the tangent meets the x-axis at $$T\!\left(\dfrac{4}{\sqrt{3}},\,0\right).$$

For the normal $$y=\dfrac{x}{\sqrt{3}}$$ we again set $$y=0$$:

$$0=\frac{x}{\sqrt{3}}\quad\Longrightarrow\quad x=0.$$

So the normal meets the x-axis at the origin $$O(0,0).$$

Therefore the triangle in question has vertices $$O(0,0),\;T\!\left(\dfrac{4}{\sqrt{3}},\,0\right),\;P\!\left(\sqrt{3},\,1\right).$$

The base $$OT$$ lies on the x-axis. Its length is the absolute difference of the x-coordinates:

$$OT=\frac{4}{\sqrt{3}}-0=\frac{4}{\sqrt{3}}.$$

The height of the triangle is the perpendicular distance from the vertex $$P$$ to the x-axis. Because the x-axis is the line $$y=0$$, that distance is simply the y-coordinate of $$P$$, namely $$1$$.

Using the formula “area = $$\dfrac12\times\text{base}\times\text{height}$$”, we substitute:

$$\text{Area}=\frac12\left(\frac{4}{\sqrt{3}}\right)(1)=\frac{2}{\sqrt{3}}.$$

Hence, the correct answer is Option B.

We begin with the two given adjacent vertices of the rectangle, $$A(-8,5)$$ and $$B(6,5)$$. Since the ordinates are equal, the segment $$AB$$ is horizontal.

Distance formula: for points $$(x_1,y_1)$$ and $$(x_2,y_2)$$ the distance is $$\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}.$$ Substituting $$A(-8,5)$$ and $$B(6,5)$$ we get $$AB=\sqrt{(6-(-8))^2+(5-5)^2}=\sqrt{14^2+0}=14.$$ So the base of the rectangle is $$14$$ units long.

Let the unknown height of the rectangle be $$h>0$$, but we do not yet decide whether the other two vertices lie above or below the segment $$AB$$. Because the sides of a rectangle are perpendicular, the two remaining vertices will share the same abscissae as $$A$$ and $$B$$ and their ordinates will differ from 5 by $$h$$.

If the rectangle rises above $$AB$$ we have $$C(6,5+h),\;D(-8,5+h).$$ If it drops below $$AB$$ we have $$C(6,5-h),\;D(-8,5-h).$$

The centre $$O$$ of a rectangle circumscribed by a circle is the intersection point of its diagonals, i.e. the midpoint of any diagonal. Using the midpoint formula $$\text{Midpoint}=\Bigl(\dfrac{x_1+x_2}{2},\dfrac{y_1+y_2}{2}\Bigr),$$ we take diagonal $$AC$$ in each case.

Case 1 (vertices above): $$O=\Bigl(\dfrac{-8+6}{2},\dfrac{5+(5+h)}{2}\Bigr)=\Bigl(-1,\dfrac{10+h}{2}\Bigr).$$

Case 2 (vertices below): $$O=\Bigl(\dfrac{-8+6}{2},\dfrac{5+(5-h)}{2}\Bigr)=\Bigl(-1,\dfrac{10-h}{2}\Bigr).$$

We are told that a diameter of the circle lies along the straight line $$3y=x+7$$. A diameter’s every point, and therefore the centre of the circle itself, must satisfy the line’s equation.

So we substitute the centre’s coordinates into $$3y=x+7$$ and solve for $$h$$.

For Case 1: $$3\Bigl(\dfrac{10+h}{2}\Bigr)=-1+7=6.$$ Hence $$\dfrac{30+3h}{2}=6\;\Longrightarrow\;30+3h=12\;\Longrightarrow\;3h=-18\;\Longrightarrow\;h=-6.$$ But $$h$$, being a length, cannot be negative, so this case is impossible.

For Case 2: $$3\Bigl(\dfrac{10-h}{2}\Bigr)=6.$$ Thus $$\dfrac{30-3h}{2}=6\;\Longrightarrow\;30-3h=12\;\Longrightarrow\;3h=18\;\Longrightarrow\;h=6.$$ Here $$h$$ is positive, so this configuration is valid: the rectangle extends 6 units below the segment $$AB$$.

Finally, the area of the rectangle equals base × height: $$\text{Area}=AB\cdot h=14\times6=84\ \text{square units}.$$

Hence, the correct answer is Option D.

We have the circle $$x^2 + y^2 - 6x + 8y - 103 = 0$$. To get its centre and radius, we convert it into standard form by completing the squares.

First group the $$x$$-terms and $$y$$-terms: $$x^2 - 6x + y^2 + 8y = 103.$$

Now complete the square for each variable. For $$x^2 - 6x$$, we add and subtract $$\left(\tfrac{6}{2}\right)^2 = 9$$. For $$y^2 + 8y$$, we add and subtract $$\left(\tfrac{8}{2}\right)^2 = 16$$. Hence

$$x^2 - 6x + 9 + y^2 + 8y + 16 = 103 + 9 + 16.$$

Simplifying, we get $$(x - 3)^2 + (y + 4)^2 = 128.$$

So the centre of the circle is $$C(3,\,-4)$$ and the radius is $$R = \sqrt{128} = 8\sqrt{2}.$$

An axis-parallel square inscribed in a circle and sharing the same centre has its four vertices at $$(h \pm a,\; k \pm a),$$ where $$(h,k)$$ is the centre and $$a$$ is the distance from the centre to the midpoint of any side (i.e. half the side length).

The distance from the centre to any vertex is given by the distance formula $$\text{Distance} = \sqrt{(\pm a)^2 + (\pm a)^2} = \sqrt{2a^2} = a\sqrt{2}.$$

Because each vertex lies on the circle, this distance must equal the radius $$R$$. Therefore $$a\sqrt{2} = 8\sqrt{2}\;\; \Longrightarrow \;\; a = 8.$$

Thus the four vertices of the square are obtained by substituting $$h = 3,\; k = -4,\; a = 8$$: $$\begin{aligned} V_1 &: (3 + 8,\; -4 + 8) = (11,\; 4),\\ V_2 &: (3 + 8,\; -4 - 8) = (11,\; -12),\\ V_3 &: (3 - 8,\; -4 + 8) = (-5,\; 4),\\ V_4 &: (3 - 8,\; -4 - 8) = (-5,\; -12). \end{aligned}$$

We must now find which vertex is closest to the origin $$(0,0)$$. Using the distance formula $$d = \sqrt{x^2 + y^2}$$ for each vertex:

$$\begin{aligned} d(V_1) &= \sqrt{11^2 + 4^2} = \sqrt{121 + 16} = \sqrt{137},\\[2pt] d(V_2) &= \sqrt{11^2 + (-12)^2} = \sqrt{121 + 144} = \sqrt{265},\\[2pt] d(V_3) &= \sqrt{(-5)^2 + 4^2} = \sqrt{25 + 16} = \sqrt{41},\\[2pt] d(V_4) &= \sqrt{(-5)^2 + (-12)^2} = \sqrt{25 + 144} = \sqrt{169} = 13. \end{aligned}$$

The smallest of these distances is $$\sqrt{41}$$, which occurs at the vertex $$(-5,\,4)$$.

Hence, the correct answer is Option C.

Let the circle of fixed radius $$R$$ pass through the origin $$O(0,0)$$ and cut the coordinate axes at the points $$A(a,0)$$ on the $$x$$-axis and $$B(0,b)$$ on the $$y$$-axis. Because $$A$$ and $$B$$ lie on the axes, the straight line joining them has intercept form

$$\frac{x}{a}+\frac{y}{b}=1.$$

We first rewrite this line in the form $$lx+my+n=0$$, because the formulas for the foot of a perpendicular from a point to a line are most convenient in that form. Multiplying by $$ab$$ gives

$$bx+ay-ab=0.$$

Thus $$l=b,\;m=a,\;n=-ab.$$

The foot of the perpendicular drawn from the origin to a line $$lx+my+n=0$$ is given by the standard formula

$$\bigl(h,k\bigr)=\left(\!-\frac{ln}{l^{2}+m^{2}},-\frac{mn}{l^{2}+m^{2}}\!\right).$$

Substituting $$l=b,\;m=a,\;n=-ab$$ we obtain

$$h=-\frac{b(-ab)}{b^{2}+a^{2}}=\frac{ab^{2}}{a^{2}+b^{2}},\qquad k=-\frac{a(-ab)}{b^{2}+a^{2}}=\frac{a^{2}b}{a^{2}+b^{2}}.$$

Hence, if $$P(h,k)$$ is the foot of the perpendicular from $$O$$ to $$AB$$, its coordinates are

$$x=h=\frac{ab^{2}}{a^{2}+b^{2}},\qquad y=k=\frac{a^{2}b}{a^{2}+b^{2}}.$$

Next we employ the fact that $$O,\,A,\,B$$ lie on the given circle of radius $$R$$. Since $$OA$$ is along the $$x$$-axis and $$OB$$ is along the $$y$$-axis, the angle at $$O$$ is a right angle. For a right-angled triangle, the circumradius equals half the hypotenuse. Therefore, using the hypotenuse $$AB,$$ we have

$$R=\frac{1}{2}\,AB=\frac{1}{2}\sqrt{a^{2}+b^{2}}\;,$$

so

$$a^{2}+b^{2}=(2R)^{2}=4R^{2}.$$

To obtain the required locus, we eliminate $$a$$ and $$b$$ from the expressions for $$x$$ and $$y$$. First compute $$x^{2}+y^{2}:$$

$$\begin{aligned} x^{2}+y^{2}&=\left(\frac{ab^{2}}{a^{2}+b^{2}}\right)^{\!2}+\left(\frac{a^{2}b}{a^{2}+b^{2}}\right)^{\!2} \\ &=\frac{a^{2}b^{4}+a^{4}b^{2}}{(a^{2}+b^{2})^{2}}\\ &=\frac{a^{2}b^{2}(b^{2}+a^{2})}{(a^{2}+b^{2})^{2}}\\ &=\frac{a^{2}b^{2}}{a^{2}+b^{2}}\\[4pt] &=\frac{a^{2}b^{2}}{4R^{2}}\quad\bigl(\text{because }a^{2}+b^{2}=4R^{2}\bigr). \end{aligned}$$

Now compute $$x^{2}y^{2}:$$

$$x^{2}y^{2}=\left(\frac{ab^{2}}{a^{2}+b^{2}}\right)^{\!2}\!\left(\frac{a^{2}b}{a^{2}+b^{2}}\right)^{\!2} =\frac{a^{6}b^{6}}{(a^{2}+b^{2})^{4}} =\frac{a^{6}b^{6}}{(4R^{2})^{4}} =\frac{a^{6}b^{6}}{256R^{8}}.$$

We are now ready to relate $$x^{2}+y^{2}$$ to $$x^{2}y^{2}$$. Cubing the earlier expression for $$x^{2}+y^{2}$$ gives

$$\bigl(x^{2}+y^{2}\bigr)^{3} =\left(\frac{a^{2}b^{2}}{4R^{2}}\right)^{\!3} =\frac{a^{6}b^{6}}{64R^{6}}.$$

On the other hand, multiplying $$x^{2}y^{2}$$ by $$4R^{2}$$ yields

$$4R^{2}x^{2}y^{2}=4R^{2}\cdot\frac{a^{6}b^{6}}{256R^{8}} =\frac{a^{6}b^{6}}{64R^{6}}.$$

Both expressions are identical, so we have the relation

$$\bigl(x^{2}+y^{2}\bigr)^{3}=4R^{2}x^{2}y^{2}.$$

This is precisely the equation listed in Option B.

Hence, the correct answer is Option 2.

We begin by noting that the given circle is $$x^2 + y^2 = 1$$ whose centre is the origin $$O(0,0)$$ and whose radius is $$1.$$

If a straight line cuts the $$x$$-axis at the point $$P(a,0)$$ and the $$y$$-axis at the point $$Q(0,b)$$, then its equation in the intercept form is

$$\frac{x}{a} + \frac{y}{b} = 1.$$

This line is stated to be a tangent to the circle. A standard result for tangency is used here:

Formula: The distance of the centre $$(x_0,y_0)$$ of a circle from a line $$\alpha x + \beta y + \gamma = 0$$ must equal the radius $$r$$ of the circle for the line to be tangent. The distance is

$$\text{Distance} = \frac{|\alpha x_0 + \beta y_0 + \gamma|}{\sqrt{\alpha^2 + \beta^2}}.$$

For our line $$\dfrac{x}{a} + \dfrac{y}{b} - 1 = 0$$ we have $$\alpha = \dfrac{1}{a},\; \beta = \dfrac{1}{b},\; \gamma = -1.$$ The centre is $$O(0,0),$$ so substituting into the distance formula we obtain

$$\frac{\left|\dfrac{1}{a}\cdot 0 + \dfrac{1}{b}\cdot 0 - 1\right|}{\sqrt{\left(\dfrac{1}{a}\right)^2 + \left(\dfrac{1}{b}\right)^2}} \;=\; 1.$$

Simplifying the numerator gives $$| -1 | = 1,$$ so the condition reduces to

$$\frac{1}{\sqrt{\dfrac{1}{a^{2}} + \dfrac{1}{b^{2}}}} = 1.$$

Taking reciprocals on both sides yields

$$\sqrt{\frac{1}{a^{2}} + \frac{1}{b^{2}}} = 1.$$

Squaring both sides we get

$$\frac{1}{a^{2}} + \frac{1}{b^{2}} = 1. \quad -(1)$$

Now we focus on the mid-point $$M(h,k)$$ of the chord $$PQ$$ connecting the intercepts. Since $$P(a,0)$$ and $$Q(0,b),$$ the mid-point coordinates are obtained via the two-point midpoint formula:

Formula: For points $$(x_1,y_1)$$ and $$(x_2,y_2),$$ their midpoint is $$\left(\dfrac{x_1+x_2}{2},\dfrac{y_1+y_2}{2}\right).$$

Applying this, we have

$$h = \frac{a + 0}{2} = \frac{a}{2}, \qquad k = \frac{0 + b}{2} = \frac{b}{2}.$$

From these relations we can express the intercepts in terms of the midpoint coordinates:

$$a = 2h, \qquad b = 2k.$$ \quad -(2)

Substituting (2) into the tangency condition (1), we obtain

$$\frac{1}{(2h)^2} + \frac{1}{(2k)^2} = 1.$$

Recognising $$(2h)^2 = 4h^2$$ and $$(2k)^2 = 4k^2,$$ the equation becomes

$$\frac{1}{4h^{2}} + \frac{1}{4k^{2}} = 1.$$

Multiplying every term by $$4$$ to clear denominators, we get

$$\frac{1}{h^{2}} + \frac{1}{k^{2}} = 4.$$

To eliminate the remaining denominators, we multiply throughout by $$h^{2}k^{2}:$$

$$k^{2} + h^{2} = 4h^{2}k^{2}.$$

Re-ordering terms brings us to

$$h^{2} + k^{2} - 4h^{2}k^{2} = 0.$$

Finally, replacing the placeholder symbols $$(h,k)$$ of the midpoint by the general coordinates $$(x,y)$$ of the locus, we write

$$x^{2} + y^{2} - 4x^{2}y^{2} = 0.$$

This exactly matches Option B.

Hence, the correct answer is Option B.

Let the centres of the two circles be $$O_1$$ and $$O_2$$ with radii

$$O_1P = r_1 = 5\ \text{cm},\qquad O_2P = r_2 = 12\ \text{cm}.$$

At the point of intersection $$P$$ the angle of intersection of the circles is given to be $$90^\circ$$. The angle of intersection is defined as the angle between the tangents drawn to the two circles at the common point, and this angle is equal to the angle between the radii drawn to that point because a radius is always perpendicular to the tangent at its end point on the circle. Hence

$$\angle O_1 P O_2 = 90^\circ.$$

Thus the triangle $$\triangle O_1 P O_2$$ is a right-angled triangle with the right angle at $$P$$. Therefore, by the Pythagoras theorem, the distance $$d$$ between the two centres is

$$d = O_1O_2 = \sqrt{O_1P^2 + O_2P^2} = \sqrt{5^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13\ \text{cm}.$$

The two circles intersect in two points. Denote the second intersection point by $$Q$$. The segment $$PQ$$ is their common chord. It is well known (and can be proved from symmetry) that the common chord is perpendicular to the line joining the two centres, and its midpoint $$M$$ lies on $$\overline{O_1O_2}$$. Let

$$O_1M = x,\qquad O_2M = d - x = 13 - x.$$

In the right triangle $$\triangle O_1MP$$ we have

$$O_1P = 5,\quad O_1M = x,\quad MP = \sqrt{O_1P^2 - O_1M^2} = \sqrt{5^2 - x^2} = \sqrt{25 - x^2}.$$

Similarly, in the right triangle $$\triangle O_2MP$$ we have

$$O_2P = 12,\quad O_2M = 13 - x,\quad MP = \sqrt{O_2P^2 - O_2M^2} = \sqrt{12^2 - (13 - x)^2} = \sqrt{144 - (13 - x)^2}.$$

Because both expressions represent the same length $$MP$$, we equate them:

$$\sqrt{25 - x^2} = \sqrt{144 - (13 - x)^2}.$$

Squaring both sides gives

$$25 - x^2 = 144 - (13 - x)^2.$$

First expand the square on the right:

$$(13 - x)^2 = 13^2 - 2\cdot13\cdot x + x^2 = 169 - 26x + x^2.$$

Substituting this, we obtain

$$25 - x^2 = 144 - \bigl(169 - 26x + x^2\bigr).$$

Remove the parentheses:

$$25 - x^2 = 144 - 169 + 26x - x^2.$$

Notice that $$-x^2$$ appears on both sides; adding $$x^2$$ to each side eliminates it:

$$25 = -25 + 26x.$$

Now isolate $$x$$:

$$26x = 25 + 25 = 50 \quad\Longrightarrow\quad x = \frac{50}{26} = \frac{25}{13}\ \text{cm}.$$

With $$x$$ known, find the half-length of the chord:

$$MP = \sqrt{25 - x^2} = \sqrt{25 - \left(\frac{25}{13}\right)^2} = \sqrt{25 - \frac{625}{169}} = \sqrt{\frac{25\cdot169 - 625}{169}} = \sqrt{\frac{4225 - 625}{169}} = \sqrt{\frac{3600}{169}} = \frac{60}{13}\ \text{cm}.$$

Therefore the full length of the common chord is

$$PQ = 2\,MP = 2\left(\frac{60}{13}\right) = \frac{120}{13}\ \text{cm}.$$

Hence, the correct answer is Option A.

We begin by naming the two given circles

$$S_1 : x^2 + y^2 + 5Kx + 2y + K = 0,$$

$$S_2 : 2(x^2 + y^2) + 2Kx + 3y - 1 = 0.$$

Their common points P and Q satisfy both $$S_1 = 0$$ and $$S_2 = 0$$ simultaneously. A standard fact from coordinate geometry states:

“Every equation of the form $$S_1 + \lambda S_2 = 0$$ also passes through the same intersection points, whatever real number $$\lambda$$ we choose.”

We need a straight line (first-degree equation) through those points. Therefore we must pick $$\lambda$$ so that every second-degree term ($$x^2$$ and $$y^2$$) cancels out in $$S_1 + \lambda S_2$$.

First expand $$S_2$$ completely:

$$S_2 = 2x^2 + 2y^2 + 2Kx + 3y - 1.$$

Now write

$$S_1 + \lambda S_2 = 0.$$

Look at the $$x^2$$-term:

$$1 + 2\lambda = 0 \;\;\Longrightarrow\;\; \lambda = -\dfrac12.$$

Check that the $$y^2$$-term gives the same condition:

$$1 + 2\lambda = 1 + 2\!\left(-\dfrac12\right) = 0,$$

so $$x^2$$ and $$y^2$$ indeed vanish together. Substitute $$\lambda = -\dfrac12$$ in $$S_1 + \lambda S_2$$ and simplify every coefficient.

Coefficient of $$x$$:

$$5K + 2\lambda K = 5K + 2\!\left(-\dfrac12\right)K = 5K - K = 4K.$$ Hence the $$x$$-term is $$4Kx$$.

Coefficient of $$y$$:

$$2 + 3\lambda = 2 + 3\!\left(-\dfrac12\right) = 2 - \dfrac32 = \dfrac12.$$ Hence the $$y$$-term is $$\dfrac12\,y.$$

Constant term:

$$K + \lambda(-1) = K - \left(-\dfrac12\right) = K + \dfrac12.$$ So the constant part is $$K + \dfrac12.$$

Putting these three results together we get the radical axis (the required straight line through P and Q):

$$4Kx + \dfrac12\,y + \left(K + \dfrac12\right) = 0.$$

The problem says that the line $$L : 4x + 5y - K = 0$$ passes through the same two points. Thus the two linear equations must represent the same line. Equality of lines means that every coefficient is proportional by one common non-zero factor. Let that factor be $$\mu.$$ Write the proportionality:

$$4x + 5y - K = \mu\!\left(4Kx + \dfrac12\,y + K + \dfrac12\right).$$

Now compare coefficients one by one.

From the $$x$$-terms:

$$4 = \mu\,(4K) \;\;\Longrightarrow\;\; \mu = \dfrac4{4K} = \dfrac1K.$$

From the $$y$$-terms:

$$5 = \mu\!\left(\dfrac12\right) = \dfrac1K\!\left(\dfrac12\right) = \dfrac1{2K}.$$

Multiply both sides by $$2K$$:

$$10K = 1 \;\;\Longrightarrow\;\; K = \dfrac1{10}.$$

Finally check the constant terms. They must be in the same ratio $$\mu = \dfrac1K.$$ Hence

$$-K = \mu\!\left(K + \dfrac12\right) = \dfrac1K\!\left(K + \dfrac12\right).$$

Multiply both sides by $$K$$ to clear the denominator:

$$-K^2 = K + \dfrac12.$$

Bring everything to the left:

$$-K^2 - K - \dfrac12 = 0 \;\;\Longrightarrow\;\; K^2 + K + \dfrac12 = 0.$$

The discriminant of this quadratic is

$$\Delta = 1^2 - 4\!\left(1\right)\!\left(\dfrac12\right) = 1 - 2 = -1 \lt 0.$$

A negative discriminant means no real value of $$K$$ satisfies the constant-term condition. Consequently there is no real $$K$$ for which all three sets of coefficients can be proportional at the same time.

Therefore the given straight line cannot pass through both points of intersection of the two circles for any real $$K$$.

Hence, the correct answer is Option B.

Let us denote the required variable circle by $$S$$.

We write its centre as $$C(h,k)$$ and its radius as $$r$$. Because the locus is required only in the first quadrant, we shall always have $$h \gt 0$$ and $$k \gt 0$$.

First condition: “the circle touches the $$y$$-axis.” The $$y$$-axis is the line $$x = 0$$, and the (perpendicular) distance of the centre $$C(h,k)$$ from this line is simply $$|h|$$. Touching the $$y$$-axis means that this distance equals the radius. Hence

$$r = h \quad\text{(because }h\gt 0\text{)}.$$

Second condition: “the circle touches the circle $$x^{2}+y^{2}=1$$ externally.” The given circle has centre $$O(0,0)$$ and radius $$1$$. For two circles to touch externally, the distance between their centres equals the sum of their radii. So, by the distance formula, we must have

$$\sqrt{(h-0)^{2}+(k-0)^{2}} \;=\; 1 + r.$$

Simplifying the left‐hand side and now substituting the value $$r = h$$ from the first condition, we get

$$\sqrt{h^{2}+k^{2}} \;=\; 1 + h.$$

To remove the square root we square both sides:

$$h^{2}+k^{2} \;=\; (1+h)^{2}.$$

Expanding the right‐hand side gives

$$h^{2}+k^{2} \;=\; 1 + 2h + h^{2}.$$

We observe that the term $$h^{2}$$ occurs on both sides; subtracting $$h^{2}$$ from each side leaves

$$k^{2} \;=\; 1 + 2h.$$

Replacing $$h$$ by $$x$$ and $$k$$ by $$y$$—because every point $$C(h,k)$$ on the locus is an ordinary point $$(x,y)$$ in the coordinate plane—we obtain the Cartesian equation of the locus:

$$y^{2} \;=\; 1 + 2x.$$

Finally, since the centres lie in the first quadrant, we keep the positive square root of $$y^{2}$$ and also insist on $$x \ge 0$$. Therefore

$$y \;=\; \sqrt{\,1 + 2x\,}, \quad x \ge 0.$$

Hence, the correct answer is Option A.

We have the circle $$x^{2}+y^{2}=16$$ whose centre is clearly at the origin $$O(0,0)$$ and whose radius is $$R=4$$ because, in the general form $$x^{2}+y^{2}=R^{2}$$, the constant term is the square of the radius.

For every natural number $$n \in \mathbb N$$ we consider the straight line $$x+y=n$$. A straight line intersects the circle and gives a real chord only when the perpendicular distance of that line from the centre is not greater than the radius.

The perpendicular distance $$d$$ of the line $$x+y=n$$ from the origin is obtained from the distance formula

$$ d=\frac{|\,Ax_{0}+By_{0}+C\,|}{\sqrt{A^{2}+B^{2}}} $$

where $$A=1,\;B=1,\;C=-n$$ and $$\bigl(x_{0},y_{0}\bigr)=(0,0)$$. Substituting we get

$$ d=\frac{|\,0+0-n\,|}{\sqrt{1^{2}+1^{2}}}=\frac{n}{\sqrt{2}}. $$

For the line to cut the circle we need $$d\le R$$, that is

$$ \frac{n}{\sqrt{2}}\;\le\;4 \;\;\Longrightarrow\;\; n\le4\sqrt2\approx5.657. $$

Since $$n$$ must be a natural number, the admissible values are

$$ n=1,\,2,\,3,\,4,\,5. $$

Thus we have exactly five chords to consider.

Now, for a circle of radius $$R$$ and a chord at a perpendicular distance $$d$$ from the centre, the length $$\ell$$ of the chord is given by the standard formula

$$ \ell = 2\sqrt{R^{2}-d^{2}}. $$

Stating the formula first and then substituting, we put $$R=4$$ and $$d=\dfrac{n}{\sqrt{2}}$$ to obtain

$$\ell_{n}=2\sqrt{\,16-\left(\frac{n}{\sqrt{2}}\right)^{2}} =2\sqrt{\,16-\frac{n^{2}}{2}}.$$

We are asked for the sum of the squares of these lengths, so we square the above expression before adding. Squaring gives

$$\ell_{n}^{2} =\bigl(2\bigr)^{2}\Bigl(16-\frac{n^{2}}{2}\Bigr) =4\left(16-\frac{n^{2}}{2}\right) =64-2n^{2}.$$

Hence, for each natural number $$n$$ in the set $$\{1,2,3,4,5\}$$, the square of the chord length is $$64-2n^{2}$$. We sum all these values:

$$\begin{aligned} \sum_{n=1}^{5}\ell_{n}^{2} &=\sum_{n=1}^{5}\bigl(64-2n^{2}\bigr) \\ &=64+64+64+64+64 \;-\;2\bigl(1^{2}+2^{2}+3^{2}+4^{2}+5^{2}\bigr) \\ &=5\times64 \;-\;2\bigl(1+4+9+16+25\bigr) \\ &=320 \;-\;2\times55 \\ &=320-110 \\ &=210. \end{aligned}$$

Therefore the required sum is $$210$$. Numerically this matches option A.

Hence, the correct answer is Option A.

First we note that each of the three circles is tangent to the $$x$$-axis from above. Hence the centre of the circle whose radius is $$a$$ can be written as $$O_{1}(x_{1},\,a)$$, the centre of the circle whose radius is $$b$$ as $$O_{2}(x_{2},\,b)$$ and that of the circle whose radius is $$c$$ as $$O_{3}(x_{3},\,c)$$.

Because all three circles touch one another externally, the distance between any two centres equals the sum of the corresponding radii. Using the distance formula $$\text{(distance)}^{2}=(\text{difference of abscissae})^{2}+(\text{difference of ordinates})^{2},$$ we can write three separate equations.

For the pair $$O_{1}O_{2}$$ we have

$$\bigl(x_{1}-x_{2}\bigr)^{2}+\bigl(a-b\bigr)^{2}=\bigl(a+b\bigr)^{2}.$$

For $$O_{1}O_{3}$$ we have

$$\bigl(x_{1}-x_{3}\bigr)^{2}+\bigl(a-c\bigr)^{2}=\bigl(a+c\bigr)^{2}.$$

For $$O_{2}O_{3}$$ we have

$$\bigl(x_{2}-x_{3}\bigr)^{2}+\bigl(b-c\bigr)^{2}=\bigl(b+c\bigr)^{2}.$$

From the first equation we obtain

$$\bigl(x_{1}-x_{2}\bigr)^{2}=(a+b)^{2}-(b-a)^{2}=4ab,$$ so $$|x_{1}-x_{2}|=2\sqrt{ab}.$$

The second equation gives

$$\bigl(x_{1}-x_{3}\bigr)^{2}=(a+c)^{2}-(c-a)^{2}=4ac,$$ so $$|x_{1}-x_{3}|=2\sqrt{ac}.$$

The third gives

$$\bigl(x_{2}-x_{3}\bigr)^{2}=(b+c)^{2}-(c-b)^{2}=4bc,$$ so $$|x_{2}-x_{3}|=2\sqrt{bc}.$$

Because $$a<b<c,$$ the smallest circle is most naturally sandwiched between the other two on the common tangent. We therefore assume the left-to-right order

$$x_{2}\;<\;x_{1}\;<\;x_{3},$$

so that

$$x_{1}-x_{2}=2\sqrt{ab},\qquad x_{3}-x_{1}=2\sqrt{ac},\qquad x_{3}-x_{2}=2\sqrt{bc}.$$

Now we use the simple relation of collinear points on a line:

$$x_{3}-x_{2}=(x_{3}-x_{1})+(x_{1}-x_{2}).$$

Substituting the expressions just found, we get

$$2\sqrt{bc}=2\sqrt{ac}+2\sqrt{ab}.$$

Dividing every term by $$2\sqrt{abc}$$ simplifies the equation:

$$\frac{\sqrt{bc}}{\sqrt{abc}}=\frac{\sqrt{ac}}{\sqrt{abc}}+\frac{\sqrt{ab}}{\sqrt{abc}}.$$

Simplifying each fraction separately, we find

$$\frac{1}{\sqrt{a}}=\frac{1}{\sqrt{b}}+\frac{1}{\sqrt{c}}.$$

This is exactly the relation stated in Option A, while none of the other options follow from the geometry. Hence, the correct answer is Option A.

We are looking for the equation of a circle which satisfies two geometrical conditions and then checking which given point lies on that circle.

First, we are told that the circle touches the x-axis at the point $$(3,0)$$. When a circle touches a line, the point of contact is perpendicular to the radius drawn to that point. Hence, the centre of the circle must lie vertically above (or below) the touching point, at a distance equal to the radius. So, if we denote the centre by $$(h,k)$$ and the radius by $$r$$, touching the x-axis at $$(3,0)$$ immediately gives

$$h = 3 \quad\text{and}\quad k = r.$$

Thus the centre is $$(3,r)$$ and the preliminary equation of the circle may be written as

$$(x-3)^2 + (y-r)^2 = r^2.$$

Secondly, the circle is said to “make an intercept of length 8 on the y-axis.’’ The y-axis is the line $$x = 0$$. So we substitute $$x = 0$$ into the equation of the circle to find the y-coordinates where the circle meets the y-axis.

Substituting $$x = 0$$:

$$(0-3)^2 + (y-r)^2 = r^2.$$

This simplifies to

$$9 + (y-r)^2 = r^2.$$

Rearranging gives

$$(y-r)^2 = r^2 - 9.$$

We now find the two y-intersection points:

$$y - r = \pm\sqrt{\,r^2 - 9\,} \quad\Longrightarrow\quad y = r \pm \sqrt{\,r^2 - 9\,}.$$

The length of the intercept on the y-axis is the distance between these two points:

$$\bigl(r + \sqrt{r^2 - 9}\bigr) - \bigl(r - \sqrt{r^2 - 9}\bigr) = 2\sqrt{r^2 - 9}.$$

We are told that this length equals 8, so

$$2\sqrt{r^2 - 9} = 8.$$

Dividing by 2, we get

$$\sqrt{r^2 - 9} = 4.$$

Squaring both sides leads to

$$r^2 - 9 = 16,$$

and therefore

$$r^2 = 25 \quad\Longrightarrow\quad r = 5.$$

Since $$k = r,$$ we also have $$k = 5.$$ Thus the centre of the circle is $$(3,5)$$ and the full equation of the circle becomes

$$(x-3)^2 + (y-5)^2 = 25.$$

Now we check each option to see which point satisfies this equation.

Option A: $$(3,10).$$ Substituting $$x = 3,\; y = 10$$ gives $$(3-3)^2 + (10-5)^2 = 0^2 + 5^2 = 25,$$ which equals the right-hand side, so this point lies on the circle.

Option B: $$(2,3).$$ Substituting gives $$(2-3)^2 + (3-5)^2 = (-1)^2 + (-2)^2 = 1 + 4 = 5 \neq 25,$$ so this point is not on the circle.

Option C: $$(3,5).$$ Substituting gives $$(3-3)^2 + (5-5)^2 = 0 + 0 = 0 \neq 25,$$ so this point is not on the circle.

Option D: $$(1,5).$$ Substituting gives $$(1-3)^2 + (5-5)^2 = (-2)^2 + 0 = 4 \neq 25,$$ so this point is not on the circle.

Only Option A satisfies the circle’s equation. Hence, the correct answer is Option A.

We are asked to find the radius of a circle $$C$$ which

    • passes through the point $$(4,0)$$, and

    • touches the fixed circle $$x^2+y^2+4x-6y=12$$ externally at the common point $$(1,-1)$$.

First, we rewrite the fixed circle in centre-radius form. We complete squares:

$$\begin{aligned} x^2+4x + y^2-6y &= 12 \\ (x^2+4x+4) + (y^2-6y+9) &= 12+4+9 \\ (x+2)^2 + (y-3)^2 &= 25. \end{aligned}$$

So its centre is $$S(-2,\,3)$$ and its radius is $$5$$.

Let the required circle $$C$$ have centre $$P(h,k)$$ and radius $$r$$. Its equation is

$$ (x-h)^2 + (y-k)^2 = r^2. $$

Because $$(4,0)$$ lies on $$C$$, we have

$$ (4-h)^2 + (0-k)^2 = r^2. \qquad (1) $$

Because the two circles touch externally at $$(1,-1)$$, that point lies on $$C$$ as well, giving

$$ (1-h)^2 + (-1-k)^2 = r^2. \qquad (2) $$

Subtracting (2) from (1) eliminates $$r^2$$:

$$\bigl[(4-h)^2+k^2\bigr] - \bigl[(1-h)^2+(-1-k)^2\bigr] = 0.$$

We expand every term explicitly:

$$\begin{aligned} (4-h)^2 &= h^2-8h+16, \\ k^2 &= k^2, \\ (1-h)^2 &= h^2-2h+1, \\ (-1-k)^2 &= k^2+2k+1. \end{aligned}$$

Substituting these expansions we get

$$ \bigl[h^2-8h+16 + k^2\bigr] - \bigl[h^2-2h+1 + k^2+2k+1\bigr] = 0. $$

All $$h^2$$ and $$k^2$$ terms cancel, leaving

$$ -8h +16 -(-2h+2k+2) = 0 \;\Longrightarrow\; -8h +16 +2h -2k -2 = 0. $$

Simplifying,

$$ -6h -2k +14 = 0 \;\Longrightarrow\; 3h + k -7 = 0. $$

Hence

$$ k = 7 - 3h. \qquad (3) $$

Next, because the circles touch externally, the distance between their centres equals the sum of their radii. The distance formula is

$$ \text{Distance}= \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}. $$

Thus

$$ \sqrt{(h+2)^2 + (k-3)^2} = r + 5. \qquad (4) $$

We square (4) to remove the square root:

$$ (h+2)^2 + (k-3)^2 = (r+5)^2. \qquad (5) $$

Now we express everything in terms of $$h$$ using (3). First we find $$r^2$$ from condition (2):

Using (3) we have $$k = 7-3h$$, so

$$\begin{aligned} r^2 &= (1-h)^2 + (-1-k)^2 \\[2mm] &= (1-h)^2 + (3h-8)^2 \\[2mm] &= (h^2-2h+1) + (9h^2-48h+64) \\[2mm] &= 10h^2 - 50h + 65. \qquad (6) \end{aligned}$$

Next, we expand the left side of (5):

$$\begin{aligned} (h+2)^2 + (k-3)^2 &= (h+2)^2 + (4-3h)^2 \\[2mm] &= (h^2+4h+4) + (9h^2-24h+16) \\[2mm] &= 10h^2 - 20h + 20. \qquad (7) \end{aligned}$$

The right side of (5) is

$$ (r+5)^2 = r^2 + 10r + 25. $$

Equating (7) to that expression and substituting (6) for $$r^2$$, we obtain

$$ 10h^2 - 20h + 20 = (10h^2 - 50h + 65) + 10r + 25. $$

All $$10h^2$$ terms cancel, giving

$$ -20h + 20 = -50h + 90 + 10r. $$

Re-arranging,

$$ 30h - 70 = 10r \;\Longrightarrow\; r = 3h - 7. \qquad (8) $$

Equations (6) and (8) must be consistent, so we substitute $$r = 3h-7$$ into (6):

$$ (3h-7)^2 = 10h^2 - 50h + 65. $$

Expanding the left side,

$$ 9h^2 - 42h + 49 = 10h^2 - 50h + 65. $$

Bringing every term to one side,

$$ 0 = 10h^2 - 50h + 65 - 9h^2 + 42h - 49 = h^2 - 8h + 16. $$

This quadratic factors neatly:

$$ (h-4)^2 = 0 \;\Longrightarrow\; h = 4. $$

Putting $$h=4$$ into (8) gives

$$ r = 3(4) - 7 = 12 - 7 = 5. $$

Therefore, the radius of the required circle $$C$$ is $$5$$ units.

Hence, the correct answer is Option B.

The two circles are

$$x^{2}+y^{2}-2x-2y-2=0$$ and $$x^{2}+y^{2}-6x-6y+14=0.$$

Write each in centre-radius form by completing squares.

First circle:
$$(x^{2}-2x)+(y^{2}-2y)=2$$
$$(x-1)^{2}-1+(y-1)^{2}-1=2$$
$$(x-1)^{2}+(y-1)^{2}=4.$$
Hence $$C_{1}(1,1), \; r_{1}=2.$$

Second circle:
$$(x^{2}-6x)+(y^{2}-6y)=-14$$
$$(x-3)^{2}-9+(y-3)^{2}-9=-14$$
$$(x-3)^{2}+(y-3)^{2}=4.$$
Hence $$C_{2}(3,3), \; r_{2}=2.$$

Subtracting the two equations to obtain the common chord:

$$(x^{2}+y^{2}-6x-6y+14)-(x^{2}+y^{2}-2x-2y-2)=0$$
$$-4x-4y+16=0 \;\Longrightarrow\; x+y=4.$$

This line is the radical axis and therefore contains the intersection points $$P,Q.$$

Put $$y=4-x$$ in the first circle:

$$x^{2}+(4-x)^{2}-2x-2(4-x)-2=0$$
$$x^{2}+x^{2}-8x+16-2x-8+2x-2=0$$
$$2x^{2}-8x+6=0$$
$$x^{2}-4x+3=0$$
$$(x-1)(x-3)=0 \;\Longrightarrow\; x=1 \text{ or } 3.$$

Hence the intersection points are

$$P(1,3), \qquad Q(3,1).$$

The required quadrilateral is $$P\,C_{1}\,Q\,C_{2}.$$ List its vertices in order:
$$P(1,3),\; C_{1}(1,1),\; Q(3,1),\; C_{2}(3,3).$$

Observe that
$$P \rightarrow C_{1}$$ is vertical of length $$2,$$
$$C_{1} \rightarrow Q$$ is horizontal of length $$2,$$
$$Q \rightarrow C_{2}$$ is vertical of length $$2,$$
$$C_{2} \rightarrow P$$ is horizontal of length $$2.$$
Thus the quadrilateral is a rectangle of side $$2 \times 2.$$

Area of rectangle $$= 2 \times 2 = 4 \text{ square units}.$$

Therefore, the area of $$P\,C_{1}\,Q\,C_{2}$$ is $$4.$$

Answer : Option B

We have two circles.

The first circle is given by $$x^2+y^2=4$$. Comparing with the standard form $$(x-h)^2+(y-k)^2=r^2,$$ we see that its centre is $$C_1(0,0)$$ and its radius is $$r_1=2.$$

The second circle is $$x^2+y^2+6x+8y-24=0.$$ To locate its centre and radius we complete squares:

$$\begin{aligned} x^2+6x+y^2+8y &= 24 \\[4pt] (x^2+6x+9)+(y^2+8y+16) &= 24+9+16 \\[4pt] (x+3)^2+(y+4)^2 &= 49. \end{aligned}$$

Thus the centre of the second circle is $$C_2(-3,-4)$$ and its radius is $$r_2=7.$$

We now look for a straight line that is tangent to both circles. Let its equation in slope-intercept form be

$$y=mx+c.$$

For any circle, a line $$y=mx+c$$ is tangent if the perpendicular distance from the centre to the line equals the radius. The perpendicular distance from $$(x_0,y_0)$$ to $$y=mx+c$$ is

$$\frac{|y_0-mx_0-c|}{\sqrt{1+m^2}}.$$

Applying this to the first circle with centre $$C_1(0,0)$$, we get

$$\frac{|\,0-0-c|}{\sqrt{1+m^2}} = r_1 = 2,$$ so $$|c| = 2\sqrt{1+m^2}.$$

Hence $$c=\pm 2\sqrt{1+m^2}.$$ We keep the symbol $$R=\sqrt{1+m^2} \quad(\,R\gt 0\,)$$ to shorten our calculations.

Next we impose the tangency condition for the second circle. The perpendicular distance from $$C_2(-3,-4)$$ to $$y=mx+c$$ must equal $$r_2=7$$:

$$\frac{|\,(-4)-m(-3)-c|}{\sqrt{1+m^2}} = 7 \;\Longrightarrow\; |\, -4+3m-c| = 7R.$$

We consider the two possible values of $$c$$ separately.

(i) Taking $$c=2R$$

The condition becomes $$|\, -4+3m-2R| = 7R.$$ Removing the absolute value gives two equations:

$$-4+3m-2R = 7R \quad\text{or}\quad -4+3m-2R = -7R.$$

Simplifying, the first yields $$3m-9R = 4,$$ and the second yields $$3m+5R = 4.$$ Carrying out the algebra (substituting $$R=\sqrt{1+m^2}$$ and squaring to remove the square root) shows that the first relation leads to no real value of $$m$$, whereas the second produces the quadratic

$$16m^2+24m+9=0.$$

The discriminant of this quadratic is $$\Delta = 24^2-4\!\cdot\!16\!\cdot\!9 = 576-576 = 0,$$ so there is one real root:

$$m = -\frac{24}{2\!\cdot\!16} = -\frac34.$$

For this value of $$m$$ we compute $$R=\sqrt{1+m^2}=\sqrt{1+\tfrac{9}{16}}=\sqrt{\tfrac{25}{16}}=\tfrac54,$$ and with $$c=2R$$ we obtain

$$c = 2\!\left(\tfrac54\right) = \tfrac52.$$

Thus the required common tangent is

$$y = -\frac34\,x + \frac52.$$

Multiplying throughout by 4 to clear fractions, we get the convenient form

$$3x + 4y - 10 = 0.$$

(ii) Taking $$c=-2R$$

A similar examination (starting with $$|\, -4+3m+2R| = 7R$$) gives no real slope, so the line found in part (i) is the only common tangent for the two circles.

We now test which of the given points satisfies the equation $$3x+4y-10=0$$:

For $$(4,-2):\;3(4)+4(-2)-10 = 12-8-10 = -6 \neq 0.$$

For $$(-4,6):\;3(-4)+4(6)-10 = -12+24-10 = 2 \neq 0.$$

For $$(6,-2):\;3(6)+4(-2)-10 = 18-8-10 = 0.$$

For $$(-6,4):\;3(-6)+4(4)-10 = -18+16-10 = -12 \neq 0.$$

Only the point $$(6,-2)$$ satisfies the equation of the common tangent.

Hence, the correct answer is Option 3.

Let the two required circles be

$$S_1:(x-a)^2+y^2=r^2\qquad\text{and}\qquad S_2:(x-c)^2+y^2=r^2,$$

where $$r$$ is their common radius, and the centres are $$C_1(a,0)$$ and $$C_2(c,0)$$ to be found. Both circles pass through the two common points $$(0,1)$$ and $$(0,-1).$$

We first use the fact that a point $$(x_1,y_1)$$ lies on a circle centred at $$(h,k)$$ with radius $$r$$ iff

$$ (x_1-h)^2+(y_1-k)^2=r^2. $$

Applying this to $$S_1$$ with the point $$(0,1)$$ gives

$$ (0-a)^2+(1-0)^2=r^2\;\Longrightarrow\;a^2+1=r^2. \quad -(1)$$

Similarly, substituting $$(0,-1)$$ in $$S_1$$ gives

$$ (0-a)^2+(-1-0)^2=r^2\;\Longrightarrow\;a^2+1=r^2. \quad -(2)$$

Equations (1) and (2) are identical, so no new condition arises; however, notice that they force the y-coordinate of the centre to be $$0$$, confirming our choice $$C_1(a,0).$$ Repeating the same calculation for $$S_2$$ shows that its centre must also lie on the x-axis, so we may safely write it as $$C_2(c,0).$$

Next, we translate the given geometric condition into an equation. We are told that “the tangent at the point $$(0,1)$$ to one of the circles passes through the centre of the other circle.” Take the tangent to $$S_1$$ at $$(0,1).$$

The vector from $$C_1(a,0)$$ to $$(0,1)$$ is $$(-a,\,1),$$ so the slope of the radius $$C_1P$$ is

$$m_{\text{radius}}=\frac{1}{-a}=-\frac1a.$$

For a line perpendicular to this radius (i.e. the tangent), the product of slopes must be $$-1,$$ hence

$$m_{\text{tangent}}\;=\;a.$$

The tangent passes through $$(0,1),$$ so by the point-slope form we have

$$ y-1 = a(x-0)\;\Longrightarrow\;y=ax+1. \quad -(3)$$

This tangent must pass through the centre $$C_2(c,0).$$ Substituting $$(x,y)=(c,0)$$ into (3) we get

$$ 0 = ac + 1\;\Longrightarrow\;c=-\frac1a. \quad -(4)$$

Because the radii of the two circles are equal, we set their squared radii equal using (1):

$$ a^2 + 1\;=\;r^2\;=\;c^2 + 1. $$

Cancelling the 1’s on both sides gives

$$ a^2 = c^2. \quad -(5)$$

Now substitute the value of $$c$$ from (4) into (5):

$$ a^2 = \left(-\frac1a\right)^2 = \frac1{a^2}. $$

Multiplying by $$a^2$$ on both sides yields

$$ a^4 = 1\;\Longrightarrow\;a^2 = 1\;\Longrightarrow\;a = \pm1. $$

Using (4) we correspondingly get

$$ c = -\frac1a = \begin{cases} -1 & \text{if } a=1,\\[6pt] \;1 & \text{if } a=-1. \end{cases} $$

Thus in either case the two centres are $$C_1(1,0)$$ and $$C_2(-1,0)$$ or vice-versa. The distance $$d$$ between them is

$$ d = |a-c| = \bigl|\,a - \bigl(-\tfrac1a\bigr)\bigr| = |\,a + \tfrac1a\,|. $$

For $$a = \pm1,$$ this becomes

$$ d = |\,\pm1 + \tfrac1{\pm1}\,| = |\,\pm1 \pm 1\,| = 2. $$

Hence, the correct answer is Option B.