JEE 2D Coordinate Geometry Questions

We must find the area of the set

$$S=\Bigl\{(x,y)\in\mathbb{R}\times\mathbb{R}\;:\;x\gt 0,\;y\gt\frac1x,\;5x-4y-1\gt 0,\;4x+4y-17\lt 0\Bigr\}.$$

Rewrite the straight-line inequalities in the usual $$y=m x+c$$ form:

$$5x-4y-1\gt0\;\Longrightarrow\;4y\lt5x-1\;\Longrightarrow\;y&lt\frac{5x-1}{4},$$
$$4x+4y-17&lt0;;\Longrightarrow\;4y&lt17-4x\;\Longrightarrow\;y&lt\frac{17-4x}{4}=\,\frac{17}{4}-x.$$

Therefore every point of $$S$$ must satisfy

$$y&gt\frac1x \quad\text{and}\quad y&lt\min\!\Bigl\{\frac{5x-1}{4},\,\frac{17}{4}-x\Bigr\}.$$

Call the two lines

$$\ell_1:\;y=\frac{5x-1}{4}\qquad \ell_2:\;y=\frac{17}{4}-x.$$

Intersection of the two lines
Set $$\frac{5x-1}{4}=\frac{17}{4}-x$$ to obtain

$$5x-1=17-4x\;\Longrightarrow\;9x=18\;\Longrightarrow\;x=2$$
and hence $$y=\frac{5(2)-1}{4}=\frac94.$$
Thus $$\ell_1\cap\ell_2=(2,\tfrac94).$$

Which line is lower?
Solve $$\frac{5x-1}{4}\lt\frac{17}{4}-x\quad\Longrightarrow\quad x&lt2.$$
Hence
for $$0&lt x&lt2$$ the lower (controlling) upper-boundary is $$\ell_1,$$
for $$x&gt2$$ the lower upper-boundary is $$\ell_2.$$

Feasibility with the hyperbola
The region exists only where the hyperbola is below the chosen line:

1. For $$y&lt\ell_1$$ (i.e. $$0&lt x&lt2$$) we need $$\frac1x&lt\frac{5x-1}{4}\;.$$ Multiply by $$4x\;(x&gt0):\quad4&lt5x^{2}-x\;,$$
$$5x^{2}-x-4&gt0;;\Longrightarrow\;x&gt1\quad(\text{rejected root }x&lt0).$$

2. For $$y&lt\ell_2$$ (i.e. $$x&gt2$$) we need $$\frac1x&lt\frac{17}{4}-x.$$ Multiply by $$4x\;(x&gt0):\quad17x-4x^{2}-4&gt0,$$
$$-4x^{2}+17x-4&gt0;;\Longrightarrow\;4x^{2}-17x+4&lt0.$$
This quadratic has roots $$x=\frac{17\pm15}{8}\;=\;0.25,\;4,$$ so the inequality holds for $$0.25&lt x&lt4.$$ Intersecting with $$x&gt2$$ leaves the range $$2&lt x&lt4.$

Thus the region splits into two x-intervals

• $$1&lt x&lt2:\quad$$\frac$$1x&lt y&lt$$\frac{5x-1}{4}$$,$$
• $$2&lt x&lt4:\quad$$\frac$$1x&lt y&lt$$\frac{17}{4}$$-x.$$

Area computation

Area $$A=$$\int_{1}^{2}\Bigl(\frac{5x-1}{4}-\frac1x\Bigr$$)\,dx+$$\int_{2}^{4}\Bigl(\frac{17}{4}-x-\frac1x\Bigr$$)\,dx.$$

First integral:

$$$$\int_{1}^{2}\!\Bigl(\frac{5x}{4}-\frac14-\frac1x\Bigr$$)dx =\Bigl[$$\frac{5x^{2}$$}{8}\Bigr]_{1}^{2}-\Bigl[$$\frac{x}{4}$$\Bigr]_{1}^{2}-\bigl[$$\ln$$ x\bigr]_{1}^{2}$$ $$=$$\frac{5}{8}$$(4-1)-$$\frac$$14(2-1)-$$\ln$$2 =$$\frac{15}{8}-\frac$$14-$$\ln$$2 =$$\frac{13}{8}-\ln$$2.$$

Second integral:

$$$$\int_{2}^{4}\!\Bigl(\frac{17}{4}-x-\frac1x\Bigr$$)dx =\Bigl[$$\frac{17x}{4}$$\Bigr]_{2}^{4}-\Bigl[$$\frac{x^{2}$$}{2}\Bigr]_{2}^{4}-\bigl[$$\ln$$ x\bigr]_{2}^{4}$$ $$=$$\frac{17}{4}$$(4-2)-$$\frac$$12(16-4)-($$\ln$$4-$$\ln$$2)$$ $$=$$\frac{17}{2}$$-6-$$\ln$$2 =$$\frac{5}{2}-\ln$$2.$$

Adding:

$$A=\Bigl($$\frac{13}{8}-\ln$$2\Bigr)+\Bigl($$\frac{5}{2}-\ln$$2\Bigr) =$$\frac{13}{8}+\frac{20}{8}$$-2$$\ln$$2 =$$\frac{33}{8}-\ln$$4.$$

Final result

Area $$=\displaystyle$$\frac{33}{8}-\log_e4$$.$$ Hence the correct option is

Option B which is: $$\dfrac{33}{8}-$$\log_e4$$$$.

To find the relationship between the sets $$A$$ and $$B$$, we can analyze the geometrical regions they represent in the $$\alpha\beta$$-plane.

The set $$A$$ is defined by the system of inequalities:

$$|\alpha - 1| \le 4 \quad \text{and} \quad |\beta - 5| \le 6$$

Expanding these inequalities, we get:

$$-4 \le \alpha - 1 \le 4 \implies -3 \le \alpha \le 5$$

$$-6 \le \beta - 5 \le 6 \implies -1 \le \beta \le 11$$

Geometrically, set $$A$$ represents a rectangular region bounded by the intervals $$\alpha \in [-3, 5]$$ and $$\beta \in [-1, 11]$$.

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The set $$B$$ is defined by the inequality:

$$16(\alpha - 2)^2 + 9(\beta - 6)^2 \le 144$$

Dividing both sides of the equation by 144 to put it in standard form:

$$\frac{16(\alpha - 2)^2}{144} + \frac{9(\beta - 6)^2}{144} \le 1$$

$$\frac{(\alpha - 2)^2}{9} + \frac{(\beta - 6)^2}{16} \le 1$$

$$\frac{(\alpha - 2)^2}{3^2} + \frac{(\beta - 6)^2}{4^2} \le 1$$

Geometrically, set $$B$$ represents the interior and boundary of a vertical ellipse centered at $$(2, 6)$$.

The boundaries (extremes) of the elliptical region $$B$$ can be found using its semi-minor axis $$a = 3$$ and semi-major axis $$b = 4$$, from here we get :

$$\alpha_{\text{range}} = [2 - 3, 2 + 3] \implies \alpha \in [-1, 5]$$

$$\beta_{\text{range}} = [6 - 4, 6 + 4] \implies \beta \in [2, 10]$$

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Comparing the coordinate bounds of both regions:

For the horizontal coordinate: the ellipse spans $$[-1, 5]$$, which is entirely inside the rectangle's range of $$[-3, 5]$$.

For the vertical coordinate: the ellipse spans $$[2, 10]$$, which is entirely inside the rectangle's range of $$[-1, 11]$$.

Since every coordinate point $$(\alpha, \beta)$$ that satisfies the ellipse inequality also satisfies the rectangular boundary inequalities, the entire elliptical region $$B$$ fits completely within the rectangular region $$A$$.

Therefore, the correct relationship is $$B \subset A$$.

Let the position vectors of the vertices be $$\mathbf{p},\;\mathbf{q},\;\mathbf{r}$$ for the points $$P,\;Q,\;R$$ respectively, and let $$\mathbf{s}$$ be the position vector of the point $$S$$.

The given vector condition is
$$\overrightarrow{SP}+5\,\overrightarrow{SQ}+6\,\overrightarrow{SR}=\vec{0}.$$

Using $$\overrightarrow{SP}=\mathbf{p}-\mathbf{s},\;\overrightarrow{SQ}=\mathbf{q}-\mathbf{s},\;\overrightarrow{SR}=\mathbf{r}-\mathbf{s},$$ we get

$$\bigl(\mathbf{p}-\mathbf{s}\bigr)+5\bigl(\mathbf{q}-\mathbf{s}\bigr)+6\bigl(\mathbf{r}-\mathbf{s}\bigr)=\vec{0}.$$

Simplifying,
$$\mathbf{p}+5\mathbf{q}+6\mathbf{r}-12\mathbf{s}= \vec{0}\;\Longrightarrow\; \mathbf{s}= \dfrac{\mathbf{p}+5\mathbf{q}+6\mathbf{r}}{12}.$$

Next, let $$E$$ and $$F$$ be the mid-points of $$PR$$ and $$QR$$.

Mid-point position vectors:
$$\mathbf{e}= \dfrac{\mathbf{p}+\mathbf{r}}{2}, \qquad \mathbf{f}= \dfrac{\mathbf{q}+\mathbf{r}}{2}.$$

Length of $$EF$$
$$\overrightarrow{EF}= \mathbf{f}-\mathbf{e}= \dfrac{\mathbf{q}+\mathbf{r}}{2}-\dfrac{\mathbf{p}+\mathbf{r}}{2} =\dfrac{\mathbf{q}-\mathbf{p}}{2}.$$
Hence $$|EF|=\dfrac{1}{2}\,|\mathbf{q}-\mathbf{p}|.$$

Length of $$ES$$
$$\overrightarrow{ES}= \mathbf{s}-\mathbf{e} =\dfrac{\mathbf{p}+5\mathbf{q}+6\mathbf{r}}{12}-\dfrac{\mathbf{p}+\mathbf{r}}{2} =\dfrac{\mathbf{p}+5\mathbf{q}+6\mathbf{r}-6\mathbf{p}-6\mathbf{r}}{12} =\dfrac{-5\mathbf{p}+5\mathbf{q}}{12} =\dfrac{5}{12}\,(\mathbf{q}-\mathbf{p}).$$
Therefore $$|ES|=\dfrac{5}{12}\,|\mathbf{q}-\mathbf{p}|.$$

Required ratio
$$\dfrac{|EF|}{|ES|}= \dfrac{\dfrac{1}{2}\,|\mathbf{q}-\mathbf{p}|}{\dfrac{5}{12}\,|\mathbf{q}-\mathbf{p}|} =\dfrac{1}{2}\times\dfrac{12}{5} =\dfrac{6}{5}=1.2.$$

Hence, $$\dfrac{\text{length of }EF}{\text{length of }ES}=1.2,$$ which lies in the range 1.15-1.25.

Line through (4, -9) with slope m > 0: $$y + 9 = m(x - 4)$$.

x-intercept A: $$y = 0 \Rightarrow x = 4 + 9/m$$. OA = 4 + 9/m.

y-intercept B: $$x = 0 \Rightarrow y = -9 - 4m$$. OB = |−9 − 4m| = 9 + 4m (since m > 0).

Sum = OA + OB = 4 + 9/m + 9 + 4m = 13 + 9/m + 4m.

By AM-GM: $$9/m + 4m \geq 2\sqrt{36} = 12$$. Equality when 9/m = 4m, m = 3/2.

Minimum sum = 13 + 12 = 25.

The correct answer is Option (2): 25.

Let the initial side length of the equilateral triangle be $$a$$.

• Perimeter of 1st triangle = $$3a$$

• Perimeter of 2nd triangle (sides are halved) = $$\frac{3a}{2}$$

• Perimeter of 3rd triangle = $$\frac{3a}{4}$$

This creates an infinite geometric series with first term $$A_1 = 3a$$ and common ratio $$r = \frac{1}{2}$$:

$$P = \frac{A_1}{1 - r} = \frac{3a}{1 - \frac{1}{2}} = 6a \implies a = \frac{P}{6}$$

• Area of 1st triangle = $$\frac{\sqrt{3}}{4}a^2$$

• Area of 2nd triangle (sides halved means area is quartered) = $$\frac{1}{4}\left(\frac{\sqrt{3}}{4}a^2\right)$$

This creates an infinite geometric series with first term $$A_1 = \frac{\sqrt{3}}{4}a^2$$ and common ratio $$r = \frac{1}{4}$$:

$$Q = \frac{A_1}{1 - r} = \frac{\frac{\sqrt{3}}{4}a^2}{1 - \frac{1}{4}} = \frac{\frac{\sqrt{3}}{4}a^2}{\frac{3}{4}} = \frac{\sqrt{3}}{3}a^2 = \frac{a^2}{\sqrt{3}}$$

$$Q = \frac{\left(\frac{P}{6}\right)^2}{\sqrt{3}} = \frac{P^2}{36\sqrt{3}}$$

$$P^2 = 36\sqrt{3}Q$$

Two lines from origin O intersect $$3x + 4y = 12$$ at P and Q such that $$\triangle OPQ$$ is isosceles with $$\angle POQ = 90°$$ and $$OP = OQ$$.

The distance from O to the line $$3x + 4y = 12$$ is $$d = \frac{12}{\sqrt{9+16}} = \frac{12}{5}$$.

Since $$\angle POQ = 90°$$ and $$OP = OQ$$, the perpendicular from O to PQ bisects both $$PQ$$ and $$\angle POQ$$. Let M be the foot of the perpendicular from O to line PQ, so that $$OM = d = 12/5$$.

Because $$\angle POM = \angle QOM = 45°$$, we have
$$OP = \frac{OM}{\cos 45°} = \frac{12/5}{1/\sqrt{2}} = \frac{12\sqrt{2}}{5},$$
$$PM = OP \sin 45° = \frac{12\sqrt{2}}{5} \cdot \frac{1}{\sqrt{2}} = \frac{12}{5},$$
and hence
$$PQ = 2 \cdot PM = \frac{24}{5}.$$

Next, we calculate
$$l = OP^2 + PQ^2 + QO^2 = 2 \cdot OP^2 + PQ^2,$$
where
$$OP^2 = QO^2 = \left(\frac{12\sqrt{2}}{5}\right)^2 = \frac{288}{25},$$
$$PQ^2 = \left(\frac{24}{5}\right)^2 = \frac{576}{25}.$$
It follows that
$$l = \frac{576}{25} + \frac{576}{25} = \frac{1152}{25} = 46.08.$$

The greatest integer less than or equal to $$l$$ is $$\lfloor 46.08 \rfloor = 46$$.

The correct answer is Option (2): 46.

Since $$A(a, -2)$$ and $$B(a, 6)$$ have the same x-coordinate, side $$AB$$ is vertical (along $$x = a$$). The perpendicular bisector of $$AB$$ is horizontal, passing through the midpoint $$(a, 2)$$, so it is the line $$y = 2$$.

The circumcenter lies on this perpendicular bisector, so:

$$\frac{a}{4} = 2 \implies a = 8$$

$$A(8, -2)$$, $$B(8, 6)$$, $$C(2, -2)$$. Circumcenter: $$(5, 2)$$.

$$AB = |6 - (-2)| = 8$$

$$AC = |8 - 2| = 6$$

$$BC = \sqrt{(8-2)^2 + (6-(-2))^2} = \sqrt{36 + 64} = \sqrt{100} = 10$$

 $$6^2 + 8^2 = 100 = 10^2$$, so this is a right triangle with the right angle at $$A$$.

For a right triangle, the circumradius equals half the hypotenuse:

$$\alpha = \frac{BC}{2} = \frac{10}{2} = 5$$

$$\beta = \frac{1}{2} \times AB \times AC = \frac{1}{2} \times 8 \times 6 = 24$$

$$\gamma = AB + AC + BC = 8 + 6 + 10 = 24$$

$$\alpha + \beta + \gamma = 5 + 24 + 24 = 53$$

The correct answer is Option (2): 53.

Orthocenter property:

$$(AP\perp BC),(BP\perp AC)$$

Find (C):

$$Slope(AP=-1\Rightarrow BC)slope(=1\Rightarrow y=x+5)$$

$$Slope(BP=-\frac{2}{3}\Rightarrow AC)slope(=\frac{3}{2}\Rightarrow y=\frac{3}{2}x-\frac{11}{2})$$

Intersection ⇒ (C(21,26))
$$\alpha+\beta=47$$

Circumcenter of (PAB):

Solve perpendicular bisectors ⇒ (h + k = -21)

Final:
$$(\alpha+\beta)+2(h+k)=47+2(-21)=5$$

The sides are $$AB: x-y+4=0$$, $$BC: 3x+5y-4=0$$, and $$AC: x+y-2=0$$.

Shifting a line $$ax+by+c=0$$ inward by $$d$$ units: $$\frac{ax+by+c}{\sqrt{a^2+b^2}} = \pm d$$.

For $$AC (x+y-2=0)$$, the distance from origin is $$\frac{|-2|}{\sqrt{2}} = \sqrt{2}$$. Shifting it 1 unit "inward" (towards the triangle's centroid) results in $$x+y-(2-\sqrt{2})=0$$.

This line is closest to the origin because the original distance was $$\sqrt{2}$$ and we moved 1 unit closer.

Answer: C ($$x+y-(2-\sqrt{2})=0$$)

The line segment joining $$(5, 2)$$ and $$(2, a)$$ subtends an angle $$\frac{\pi}{4}$$ at the origin.

Let $$O = (0,0)$$, $$P = (5, 2)$$, $$Q = (2, a)$$.

$$\cos\frac{\pi}{4} = \frac{\vec{OP} \cdot \vec{OQ}}{|\vec{OP}||\vec{OQ}|}$$

$$\frac{1}{\sqrt{2}} = \frac{5(2) + 2(a)}{\sqrt{5^2 + 2^2} \cdot \sqrt{2^2 + a^2}} = \frac{10 + 2a}{\sqrt{29} \cdot \sqrt{4 + a^2}}$$

$$\frac{1}{2} = \frac{(10 + 2a)^2}{29(4 + a^2)}$$

$$29(4 + a^2) = 2(10 + 2a)^2$$

$$116 + 29a^2 = 2(100 + 40a + 4a^2) = 200 + 80a + 8a^2$$

$$21a^2 - 80a - 84 = 0$$

By Vieta's formulas for $$21a^2 - 80a - 84 = 0$$:

Sum of roots = $$\frac{80}{21}$$

Product of roots = $$\frac{-84}{21} = -4$$

Discriminant = $$80^2 + 4(21)(84) = 6400 + 7056 = 13456 = 116^2$$

$$a_1 = \frac{80 + 116}{42} = \frac{196}{42} = \frac{14}{3}, \quad a_2 = \frac{80 - 116}{42} = \frac{-36}{42} = -\frac{6}{7}$$

For both values: $$10 + 2a > 0$$, confirming the angle is in the correct range.

The product of all possible values of $$a$$ is $$\frac{14}{3} \times \left(-\frac{6}{7}\right) = -4$$.

Therefore, the answer is Option D: $$\mathbf{-4}$$.

A line through $$(3, 5)$$ with intercepts on positive axes at A$$(a, 0)$$ and B$$(0, b)$$.

Line: $$\frac{x}{a} + \frac{y}{b} = 1$$. Since it passes through $$(3, 5)$$: $$\frac{3}{a} + \frac{5}{b} = 1$$.

Area of triangle OAB = $$\frac{1}{2}ab$$.

We need to minimize $$ab$$ subject to $$\frac{3}{a} + \frac{5}{b} = 1$$ with $$a > 3, b > 5$$ (positive intercepts with point (3,5) between them).

By AM-GM: $$1 = \frac{3}{a} + \frac{5}{b} \geq 2\sqrt{\frac{15}{ab}}$$.

$$\frac{1}{4} \geq \frac{15}{ab} \implies ab \geq 60$$.

Minimum area = $$\frac{1}{2} \times 60 = 30$$.

Equality when $$\frac{3}{a} = \frac{5}{b}$$, i.e., $$a = 6, b = 10$$.

Let inner rectangle sides be $$4$$ and $$2$$, inclined at an angle $$\theta$$ to the outer rectangle sides.

Outer dimensions: $$a = 4\cos\theta + 2\sin\theta$$ and $$b = 4\sin\theta + 2\cos\theta$$.

The outer area is maximized when it forms a symmetric configuration around the inner block ($$\theta = 45^\circ$$).

Substituting $$\theta = 45^\circ$$: $$a = b = \frac{4}{\sqrt{2}} + \frac{2}{\sqrt{2}} = 3\sqrt{2}$$.

 $$(a + b)^2 = (3\sqrt{2} + 3\sqrt{2})^2 = (6\sqrt{2})^2 = 72$$.

Given points $$P(3, 2, 3)$$, $$Q(4, 6, 2)$$, and $$R(7, 3, 2)$$. 

Find the direction vectors from P: $$\vec{PQ} = Q - P = (4-3, 6-2, 2-3) = (1, 4, -1)$$

$$\vec{PR} = R - P = (7-3, 3-2, 2-3) = (4, 1, -1)$$

Compute the dot product and magnitudes: $$\vec{PQ} \cdot \vec{PR} = (1)(4) + (4)(1) + (-1)(-1) = 4 + 4 + 1 = 9$$

$$|\vec{PQ}| = \sqrt{1 + 16 + 1} = \sqrt{18} = 3\sqrt{2}$$

$$|\vec{PR}| = \sqrt{16 + 1 + 1} = \sqrt{18} = 3\sqrt{2}$$

Apply the angle formula: $$ \cos(\angle QPR) = \frac{\vec{PQ} \cdot \vec{PR}}{|\vec{PQ}||\vec{PR}|} = \frac{9}{3\sqrt{2} \cdot 3\sqrt{2}} = \frac{9}{18} = \frac{1}{2} $$

$$ \angle QPR = \cos^{-1}\left(\frac{1}{2}\right) = \frac{\pi}{3} $$

The correct answer is Option (4): $$\frac{\pi}{3}$$.

For $$x$$ to be real, the discriminant $$D \ge 0$$:

$$[2b(a+c)]^2 - 4(a^2 + b^2)(b^2 + c^2) \ge 0$$

Expanding and simplifying:

$$b^2(a^2 + c^2 + 2ac) - (a^2b^2 + a^2c^2 + b^4 + b^2c^2) \ge 0$$

$$2ab^2c - a^2c^2 - b^4 \ge 0 \implies -(b^2 - ac)^2 \ge 0$$

This is only possible if $$b^2 - ac = 0$$, meaning $$b^2 = ac$$ (Geometric Progression).

If $$D=0$$, $$x = \frac{2b(a+c)}{2(a^2+b^2)} = \frac{b(a+c)}{a^2+ac} = \frac{b(a+c)}{a(a+c)} = \frac{b}{a}$$.

Since $$b^2=ac$$, we can also write $$x = \frac{\sqrt{ac}}{a} = \sqrt{\frac{c}{a}}$$.

For a triangle: $$a+b > c$$, $$b+c > a$$, and $$a+c > b$$.

Using $$b = \sqrt{ac}$$, the set of $$x$$ (where $$x = b/a = \sqrt{c/a}$$) must satisfy:

$$1 + x > x^2$$, $$x + x^2 > 1$$, and $$1 + x^2 > x$$.

Solving these gives $$x \in (\frac{\sqrt{5}-1}{2}, \frac{\sqrt{5}+1}{2})$$.

So $$\alpha = \frac{\sqrt{5}-1}{2}$$ and $$\beta = \frac{\sqrt{5}+1}{2}$$.

$$\alpha^2 + \beta^2 = (\frac{6-2\sqrt{5}}{4}) + (\frac{6+2\sqrt{5}}{4}) = \frac{12}{4} = 3$$.

Value = $$12(3) = \mathbf{36}$$

We have an isosceles triangle ABC with $$A(-1, 0)$$, $$\angle A = \frac{2\pi}{3}$$, $$AB = AC$$, and B on the positive x-axis. Also $$BC = 4\sqrt{3}$$.

Since $$\angle A = 120°$$ and $$AB = AC$$, the base angles are each $$\frac{180° - 120°}{2} = 30°$$.

Using the sine rule: $$\frac{BC}{\sin A} = \frac{AB}{\sin C}$$

$$\frac{4\sqrt{3}}{\sin 120°} = \frac{AB}{\sin 30°}$$

$$AB = 4$$

Since B is on the positive x-axis and $$AB = 4$$, we have $$B = (-1 + 4, 0) = (3, 0)$$.

Now, since the triangle is isosceles with $$AB = AC = 4$$ and $$\angle A = 120°$$, the direction from A to C makes an angle of $$120°$$ with AB.

Direction of AB is along the positive x-axis (angle $$0°$$). So AC is at angle $$120°$$ from AB.

$$C = A + 4(\cos 120°, \sin 120°) = (-1 + 4(-\frac{1}{2}), 0 + 4 \cdot \frac{\sqrt{3}}{2}) = (-3, 2\sqrt{3})$$

Or at angle $$-120°$$: $$C = (-3, -2\sqrt{3})$$.

Let's check $$BC$$: $$BC = \sqrt{(3-(-3))^2 + (0-2\sqrt{3})^2} = \sqrt{36 + 12} = \sqrt{48} = 4\sqrt{3}$$. Correct!

Line BC passes through $$B(3, 0)$$ and $$C(-3, 2\sqrt{3})$$:

Equation: $$y = -\frac{\sqrt{3}}{3}(x - 3) = -\frac{\sqrt{3}}{3}x + \sqrt{3}$$

Intersection with $$y = x + 3$$: $$x + 3 = -\frac{\sqrt{3}}{3}x + \sqrt{3}$$

$$x = \frac{3(\sqrt{3} - 3)}{3 + \sqrt{3}} = \frac{3(\sqrt{3} - 3)(3 - \sqrt{3})}{(3 + \sqrt{3})(3 - \sqrt{3})} = \frac{3(\sqrt{3} - 3)(3 - \sqrt{3})}{9 - 3} = \frac{3(\sqrt{3} - 3)(3 - \sqrt{3})}{6}$$

$$(\sqrt{3} - 3)(3 - \sqrt{3}) = -(3 - \sqrt{3})(3 - \sqrt{3}) = -(3 - \sqrt{3})^2 = -(9 - 6\sqrt{3} + 3) = -(12 - 6\sqrt{3})$$

$$x = \frac{3 \cdot (-(12 - 6\sqrt{3}))}{6} = \frac{-3(12 - 6\sqrt{3})}{6} = \frac{-36 + 18\sqrt{3}}{6} = -6 + 3\sqrt{3}$$

$$\beta = x + 3 = -6 + 3\sqrt{3} + 3 = -3 + 3\sqrt{3} = 3(\sqrt{3} - 1)$$

$$\alpha = -6 + 3\sqrt{3} = 3(\sqrt{3} - 2)$$

For the other case $$C = (-3, -2\sqrt{3})$$, slope = $$\frac{\sqrt{3}}{3}$$, equation: $$y = \frac{\sqrt{3}}{3}(x-3) = \frac{\sqrt{3}}{3}x - \sqrt{3}$$.

Intersection: $$x + 3 = \frac{\sqrt{3}}{3}x - \sqrt{3}$$, so $$x(1 - \frac{\sqrt{3}}{3}) = -\sqrt{3} - 3$$.

$$x = \frac{-3(\sqrt{3}+3)}{3-\sqrt{3}} \cdot \frac{3+\sqrt{3}}{3+\sqrt{3}} = \frac{-3(\sqrt{3}+3)(3+\sqrt{3})}{6}$$

$$= \frac{-3(3\sqrt{3}+3+9+3\sqrt{3})}{6} = \frac{-3(12+6\sqrt{3})}{6} = -(6+3\sqrt{3})$$

$$\alpha = -6-3\sqrt{3}$$, $$\beta = -3-3\sqrt{3}$$.

Now compute $$\frac{\beta^4}{\alpha^2}$$. Using the first case:

$$\alpha = 3(\sqrt{3}-2)$$, $$\beta = 3(\sqrt{3}-1)$$

$$\frac{\beta^4}{\alpha^2} = \frac{81(\sqrt{3}-1)^4}{9(\sqrt{3}-2)^2} = \frac{9(\sqrt{3}-1)^4}{(\sqrt{3}-2)^2}$$

$$\frac{9(28-16\sqrt{3})}{7-4\sqrt{3}} = \frac{9 \cdot 4(7-4\sqrt{3})}{7-4\sqrt{3}} = 36$$

First, we find the point P in the first quadrant where the circle $$x^2 + y^2 = 3$$ intersects the parabola $$x^2 = 2y$$. 

Substituting $$x^2 = 2y$$ into the circle gives $$2y + y^2 = 3$$, so $$y^2 + 2y - 3 = 0$$ and $$(y+3)(y-1) = 0$$. 
Since P is in the first quadrant, $$y = 1$$ and $$x^2 = 2$$, hence $$x = \sqrt{2}$$, which gives $$P = (\sqrt{2},1)$$.

The line $$L: \sqrt{2}x + y = \alpha$$ passes through $$P(\sqrt{2},1)$$, so $$\sqrt{2}\cdot\sqrt{2} + 1 = 2 + 1 = 3$$, giving $$\alpha = 3$$. Therefore the equation of the line is $$\sqrt{2}x + y = 3$$.

The circles $$C_1$$ and $$C_2$$ each have radius $$2\sqrt{3}$$ and centres on the y-axis, denoted $$Q_1 = (0,k_1)$$ and $$Q_2 = (0,k_2)$$. Since line $$L$$ is tangent to each circle, the distance from $$(0,k)$$ to the line $$\sqrt{2}x + y - 3 = 0$$ must equal $$2\sqrt{3}$$. This distance is $$d = \frac{|0 + k - 3|}{\sqrt{2 + 1}} = \frac{|k - 3|}{\sqrt{3}} = 2\sqrt{3}$$, which implies $$|k - 3| = 6$$. Hence $$k = 9$$ or $$k = -3$$, giving $$Q_1 = (0,9)$$ and $$Q_2 = (0,-3)$$.

With $$P = (\sqrt{2},1)$$, $$Q_1 = (0,9)$$ and $$Q_2 = (0,-3)$$, the base $$Q_1Q_2$$ has length $$|9 - (-3)| = 12$$ and the height from P to the y-axis is $$|\sqrt{2}| = \sqrt{2}$$. Therefore the area of triangle $$PQ_1Q_2$$ is $$\frac{1}{2} \times 12 \times \sqrt{2} = 6\sqrt{2}$$.

Finally, squaring this area gives $$(6\sqrt{2})^2 = 72$$. Thus, the square of the area is 72.

We need to find point $$P$$ where lines $$ax + by + c = 0$$ are concurrent (for $$a, b, c$$ in A.P.), and point $$Q(\alpha, \beta)$$ from the system with infinitely many solutions.

Finding point P: since $$a, b, c$$ are in A.P.: $$b = a + d$$ and $$c = a + 2d$$ for some common difference $$d$$.

Substituting into $$ax + by + c = 0$$:

$$ax + (a + d)y + (a + 2d) = 0$$

$$a(x + y + 1) + d(y + 2) = 0$$

For this to hold for all values of $$a$$ and $$d$$, we need:

$$x + y + 1 = 0$$ and $$y + 2 = 0$$

So $$y = -2$$ and $$x = 1$$. Therefore $$P = (1, -2)$$.

Finding point Q: the system is: $$x + y + z = 6$$, $$2x + 5y + \alpha z = \beta$$, $$x + 2y + 3z = 4$$.

For infinitely many solutions, the determinant of the coefficient matrix must be zero:

$$\begin{vmatrix} 1 & 1 & 1 \\ 2 & 5 & \alpha \\ 1 & 2 & 3 \end{vmatrix} = 1(15 - 2\alpha) - 1(6 - \alpha) + 1(4 - 5) = 0$$

$$15 - 2\alpha - 6 + \alpha + (-1) = 0 \implies 8 - \alpha = 0 \implies \alpha = 8$$

For consistency with $$\alpha = 8$$: From equations (1) and (3): $$(3) - (1)$$: $$y + 2z = -2$$.

From equations (1) and (2): $$(2) - 2(1)$$: $$3y + 6z = \beta - 12$$, i.e., $$3(y + 2z) = \beta - 12$$.

Since $$y + 2z = -2$$: $$3(-2) = \beta - 12 \implies \beta = 6$$.

So $$Q = (\alpha, \beta) = (8, 6)$$.

Computing $$(PQ)^2$$: $$ (PQ)^2 = (8 - 1)^2 + (6 - (-2))^2 = 49 + 64 = 113 $$

The answer is $$\boxed{113}$$.

Let the centre of the circle be $$O$$ and take the complex plane with $$O$$ at the origin.
Because the radius is $$2$$, every vertex of the regular octagon can be written as $$A_k :\; 2\,\zeta^{\,k}, \qquad k = 0,1,\dots ,7$$ where $$\zeta = e^{\,i\pi/4}$$ is a primitive eighth root of unity.

Let the variable point on the circle be $$P :\; 2\,e^{\,i\theta}, \qquad 0\le \theta \lt 2\pi.$$ The distance between $$P$$ and $$A_k$$ is therefore

$$PA_k \;=\; \bigl|2e^{\,i\theta} - 2\zeta^{\,k}\bigr| = 2\,\bigl|e^{\,i\theta} - \zeta^{\,k}\bigr|.$$ Hence the required product is

$$\begin{aligned} PA_1\cdot PA_2\cdots PA_8 &= \prod_{k=0}^{7} 2\,\bigl|e^{\,i\theta}-\zeta^{\,k}\bigr| \\ &= 2^{8}\;\prod_{k=0}^{7} \bigl|e^{\,i\theta}-\zeta^{\,k}\bigr|. \end{aligned}$$

To evaluate the remaining product, recall the factorisation identity

$$\prod_{k=0}^{7}\bigl(z-\zeta^{\,k}\bigr) = z^{8}-1.$$

Taking modulus on the unit circle (i.e. $$|z|=1$$, put $$z=e^{\,i\theta}$$) gives

$$\prod_{k=0}^{7}\bigl|e^{\,i\theta}-\zeta^{\,k}\bigr| = \bigl|e^{\,i8\theta}-1\bigr|.$$

Using $$e^{\,ix}-1 = 2i\sin\dfrac{x}{2}\,e^{\,ix/2},$$ we have

$$\bigl|e^{\,i8\theta}-1\bigr| = 2\bigl|\sin 4\theta\bigr|.$$ Therefore

$$PA_1\cdot PA_2\cdots PA_8 = 2^{8}\;\cdot 2\bigl|\sin 4\theta\bigr| = 2^{9}\bigl|\sin 4\theta\bigr|.$$

The maximum value of $$|\sin 4\theta|$$ is $$1$$, so the maximum value of the whole product is

$$2^{9}\times 1 = 512.$$

Hence the required maximum value is 512.

Let the angles of the triangle be $$A \le B \le C$$ and the sides opposite to them be $$a,\,b,\,c$$ respectively.
Because the triangle is obtuse, $$C$$ is the obtuse angle. We are told that

$$C-A=\frac{\pi}{2} \qquad -(1)$$

All three vertices lie on a circle of radius $$R=1$$, so by the sine rule

$$a=2\sin A,\; b=2\sin B,\; c=2\sin C \qquad -(2)$$

The sides are in arithmetic progression, hence

$$2b=a+c \;\Longrightarrow\; 2\sin B=\sin A+\sin C \qquad -(3)$$

Using $$C=A+\frac{\pi}{2}$$ from $$(1)$$, we get $$\sin C=\sin\!\left(A+\frac{\pi}{2}\right)=\cos A$$. Also, since $$A+B+C=\pi$$,

$$B=\pi-A-C=\pi-A-\left(A+\frac{\pi}{2}\right)=\frac{\pi}{2}-2A \qquad -(4)$$

Now write every term in $$(3)$$ in terms of $$A$$:

$$\sin B=\sin\!\left(\frac{\pi}{2}-2A\right)=\cos 2A$$ $$\sin C=\cos A$$

Substituting in $$(3)$$,

$$2\cos 2A=\sin A+\cos A \qquad -(5)$$

Square both sides of $$(5)$$:

$$4\cos^2 2A=\sin^2 A+\cos^2 A+2\sin A\cos A=1+\sin 2A \qquad -(6)$$

Rewrite $$\cos^2 2A=\dfrac{1+\cos 4A}{2}$$ and put this in $$(6)$$:

$$2\bigl(1+\cos 4A\bigr)=1+\sin 2A\;\;\Longrightarrow\;\;1+2\cos 4A=\sin 2A \qquad -(7)$$

Set $$t=2A$$ (note $$0\lt t\lt \dfrac{\pi}{2}$$ because $$A\lt \dfrac{\pi}{4}$$). Equation $$(7)$$ becomes

$$1+2\cos 2t=\sin t \qquad -(8)$$

Using $$\cos 2t=1-2\sin^2 t$$, convert $$(8)$$ entirely to $$\sin t$$:

$$1+2\bigl(1-2\sin^2 t\bigr)=\sin t \;\;\Longrightarrow\;\;3-4\sin^2 t-\sin t=0$$

This is a quadratic in $$\sin t$$:

$$4\sin^2 t+\sin t-3=0 \qquad -(9)$$

Solving $$(9)$$,

$$\sin t=\frac{-1\pm\sqrt{1+48}}{8}=\frac{-1\pm7}{8}$$

The negative root is extraneous, hence

$$\sin t=\frac{3}{4} \;\;\Longrightarrow\;\; \sin 2A=\frac{3}{4} \qquad -(10)$$

Because $$0\lt 2A\lt \dfrac{\pi}{2}$$, $$\cos 2A=\sqrt{1-\left(\frac{3}{4}\right)^2}=\frac{\sqrt7}{4} \qquad -(11)$$

Half-angle formulae give

$$\sin A=\sqrt{\frac{1-\cos 2A}{2}}=\sqrt{\frac{1-\dfrac{\sqrt7}{4}}{2}} \;=\;\sqrt{\frac{4-\sqrt7}{8}}$$ $$\cos A=\sqrt{\frac{1+\cos 2A}{2}}=\sqrt{\frac{1+\dfrac{\sqrt7}{4}}{2}} \;=\;\sqrt{\frac{4+\sqrt7}{8}} \qquad -(12)$$

From $$(11)$$, $$\sin B=\cos 2A=\dfrac{\sqrt7}{4}$$ and from $$(12)$$, $$\sin C=\cos A$$.

Area of a triangle with circum-radius $$R$$ is $$\Delta=\dfrac{abc}{4R}$$. Using $$R=1$$ and $$(2)$$,

$$\Delta=2\sin A\sin B\sin C=2(\sin A)(\cos 2A)(\cos A) \qquad -(13)$$

Compute the product $$\sin A\cos A$$ from $$(12)$$:

$$\sin A\cos A=\sqrt{\frac{4-\sqrt7}{8}}\, \sqrt{\frac{4+\sqrt7}{8}} =\frac{\sqrt{(4-\sqrt7)(4+\sqrt7)}}{8} =\frac{\sqrt{16-7}}{8}=\frac{3}{8} \qquad -(14)$$

Put $$(11)$$ and $$(14)$$ in $$(13)$$:

$$\Delta=2\left(\frac{3}{8}\right)\left(\frac{\sqrt7}{4}\right) =\frac{3\sqrt7}{16} \qquad -(15)$$

Finally,

$$64\Delta = 64\left(\frac{3\sqrt7}{16}\right)=12\sqrt7$$ $$\bigl(64\Delta\bigr)^2=(12\sqrt7)^2=144\times7=1008$$

Therefore, the required value is 1008.

Let the lamp post be at point $$O$$. The person is at point $$P$$, a distance $$y$$ (in cm) from $$O$$. The tip of the shadow is at point $$T$$, so $$PT = x$$ and $$OT = y + x$$.

Heights (in cm): lamp post = $$400$$, person = $$160$$.
Because $$\triangle OTP$$ and $$\triangle PT(\text{top of person})$$ are similar,

$$\frac{400}{\,y + x\,} = \frac{160}{\,x\,}$$.

Cross-multiplying:
$$400x = 160(y + x)$$
$$400x = 160y + 160x$$
$$240x = 160y$$
$$\Rightarrow\; x = \frac{2}{3}\,y$$ $$-(1)$$

Differentiating $$(1)$$ with respect to time $$t$$,

$$\frac{dx}{dt} = \frac{2}{3}\,\frac{dy}{dt}$$ $$-(2)$$

The person walks away at $$\frac{dy}{dt} = 60\text{ cm s}^{-1}$$. Substituting in $$(2)$$:

$$\frac{dx}{dt} = \frac{2}{3}\times 60 = 40\text{ cm s}^{-1}$$.

Speed of the tip of the shadow relative to the ground:
$$\frac{d}{dt}(y + x) = \frac{dy}{dt} + \frac{dx}{dt} = 60 + 40 = 100\text{ cm s}^{-1}$$.

Required speed (tip with respect to the person) = ground speed of tip − ground speed of person:

$$100 - 60 = 40\text{ cm s}^{-1}$$.

Hence, the speed of the tip of the shadow relative to the person is 40 cm s$$^{-1}$$.

Place the triangle on the coordinate plane so that the right angle is at the origin.
Let $$A(0,0)$$, $$B(1,0)$$ and $$C(0,3)$$. Thus $$AB=1$$, $$AC=3$$ and $$\angle BAC=\frac{\pi}{2}$$.

Step 1 : Circumcircle of $$\triangle ABC$$
For a right-angled triangle the hypotenuse is the diameter of the circumcircle.
Hence the circumcircle has
center $$O$$ = midpoint of $$BC = \left(\tfrac{1}{2},\tfrac{3}{2}\right)$$ and
radius $$R = \dfrac{BC}{2} = \dfrac{\sqrt{1^{2}+3^{2}}}{2} = \dfrac{\sqrt{10}}{2} = \sqrt{2.5}$$.

Step 2 : Coordinates of the required smaller circle
The required circle touches the positive $$x$$-axis (side $$AB$$) and the positive $$y$$-axis (side $$AC$$).
If its radius is $$r$$, its centre must therefore be $$P(r,r)$$, because the distance from $$P$$ to each axis must be $$r$$.

Step 3 : Condition for internal tangency with the circumcircle
For two circles with centres $$P$$ and $$O$$ and radii $$r$$ and $$R$$, internal tangency means
$$OP + r = R$$ $$-(1)$$

First compute $$OP$$:
$$OP = \sqrt{(r-0.5)^{2} + (r-1.5)^{2}}$$.

Substitute into $$(1)$$:
$$\sqrt{(r-0.5)^{2} + (r-1.5)^{2}} + r = \sqrt{2.5}$$.

Step 4 : Solve for $$r$$

Rearrange:
$$\sqrt{(r-0.5)^{2} + (r-1.5)^{2}} = \sqrt{2.5} - r.$$ Square both sides:
$$(r-0.5)^{2} + (r-1.5)^{2} = (\sqrt{2.5} - r)^{2}.$$ Compute each side:
Left side: $$(r^{2}-r+0.25) + (r^{2}-3r+2.25) = 2r^{2} - 4r + 2.5.$$ Right side: $$2.5 - 2r\sqrt{2.5} + r^{2}.$$ Set them equal:
$$2r^{2} - 4r + 2.5 = 2.5 - 2r\sqrt{2.5} + r^{2}.$$ Simplify:
$$r^{2} - 4r + 2r\sqrt{2.5} = 0.$$ Factor $$r$$:
$$r\bigl(r - 4 + 2\sqrt{2.5}\bigr) = 0.$$ Since $$r>0$$, we obtain
$$r = 4 - 2\sqrt{2.5}.$$

Step 5 : Numerical value
$$\sqrt{2.5} \approx 1.58114 \;\;\Longrightarrow\;\; r \approx 4 - 2(1.58114) \approx 4 - 3.16228 \approx 0.83772.$$ Rounded to two decimal places, $$r \approx 0.84$$. Accepting truncation gives $$0.83$$.

Hence the admissible values are 0.83 or 0.84.

Given $$A = \left(\frac{3}{\sqrt{a}}, \sqrt{a}\right)$$, $$B$$ is the image of $$A$$ in the y-axis, and $$C$$ is the image of $$B$$ in the x-axis.

Find coordinates of B and C.

$$B = \left(-\frac{3}{\sqrt{a}}, \sqrt{a}\right)$$ (reflection of A in y-axis)

$$C = \left(-\frac{3}{\sqrt{a}}, -\sqrt{a}\right)$$ (reflection of B in x-axis)

Set up the area of triangle ACD.

$$D = (3\cos\theta, a\sin\theta)$$ is in the fourth quadrant ($$\cos\theta > 0, \sin\theta < 0$$).

Area $$= \frac{1}{2}|x_A(y_C - y_D) + x_C(y_D - y_A) + x_D(y_A - y_C)|$$

$$= \frac{1}{2}\left|\frac{3}{\sqrt{a}}(-\sqrt{a} - a\sin\theta) + \left(-\frac{3}{\sqrt{a}}\right)(a\sin\theta - \sqrt{a}) + 3\cos\theta(\sqrt{a} + \sqrt{a})\right|$$

$$= \frac{1}{2}\left|(-3 - 3\sqrt{a}\sin\theta) + (-3\sqrt{a}\sin\theta + 3) + 6\sqrt{a}\cos\theta\right|$$

$$= \frac{1}{2}\left|-6\sqrt{a}\sin\theta + 6\sqrt{a}\cos\theta\right|$$

$$= 3\sqrt{a}|\cos\theta - \sin\theta|$$

Maximize the area.

The maximum of $$|\cos\theta - \sin\theta| = \sqrt{2}$$.

Maximum area $$= 3\sqrt{a} \cdot \sqrt{2} = 12$$

$$\sqrt{a} = \frac{12}{3\sqrt{2}} = \frac{4}{\sqrt{2}} = 2\sqrt{2}$$

$$a = 8$$

Answer: 8

Let the original point be denoted by $$P(a,\,b)\,. $$ We shall follow the three prescribed transformations one after another and track the co-ordinates at every stage.

First, we perform a reflection about the line $$y=x.$$ For any point $$\bigl(x,\,y\bigr)$$, reflection in the line $$y=x$$ interchanges its co-ordinates. Hence

$$P(a, b)\;\longrightarrow\;P_1\bigl(b$$, $$\$$, $$a\bigr).$$

Next, we translate the point by 2 units along the positive $$x$$-axis. Translation along the positive $$x$$-direction by 2 units simply adds 2 to the $$x$$-coordinate while leaving the $$y$$-coordinate unchanged. Therefore

$$P_1\bigl(b,\,a\bigr)\;\longrightarrow\;P_2\bigl(b+2,\,a\bigr).$$

Now we rotate the point anticlockwise about the origin through an angle $$\dfrac{\pi}{4}\,.$$ For rotation of a point $$(x,\,y)$$ by an angle $$\theta$$ anticlockwise about the origin, the standard formula is

$$\bigl(x,\,y\bigr)\;\longrightarrow\;\Bigl(x\cos\theta-y\sin\theta,\;x\sin\theta+y\cos\theta\Bigr).$$

Here $$\theta=\dfrac{\pi}{4}$$, so $$\cos\theta=\dfrac{1}{\sqrt2}$$ and $$\sin\theta=\dfrac{1}{\sqrt2}\,.$$ Applying the formula to $$P_2\bigl(b+2,\,a\bigr)$$ we obtain the final point $$P_3(x_f,\,y_f):$$

$$\begin{aligned} x_f &= (b+2)\cos\frac{\pi}{4} - a\sin\frac{\pi}{4} = (b+2)\cdot\frac1{\sqrt2} - a\cdot\frac1{\sqrt2} = \frac{(b+2)-a}{\sqrt2},\\[4pt] y_f &= (b+2)\sin\frac{\pi}{4} + a\cos\frac{\pi}{4} = (b+2)\cdot\frac1{\sqrt2} + a\cdot\frac1{\sqrt2} = \frac{(b+2)+a}{\sqrt2}. \end{aligned}$$

We are told that the final co-ordinates are $$\left(-\dfrac1{\sqrt2},\;\dfrac7{\sqrt2}\right).$$ Hence

$$\frac{(b+2)-a}{\sqrt2}=-\frac1{\sqrt2}\quad\text{and}\quad \frac{(b+2)+a}{\sqrt2}=\frac7{\sqrt2}.$$

Multiplying both equations by $$\sqrt2$$ eliminates the denominator:

$$\begin{aligned} (b+2)-a &= -1,\\ (b+2)+a &= 7. \end{aligned}$$

Simplifying each equation gives

$$\begin{aligned} -b + a &= 3 \quad\Longrightarrow\quad a - b = 3,\\ a + b &= 5. \end{aligned}$$

Now we solve these simultaneous linear equations. Adding them yields

$$2a = 8 \;\;\Longrightarrow\;\; a = 4.$$

Substituting $$a=4$$ in $$a+b=5$$ gives

$$4 + b = 5 \;\;\Longrightarrow\;\; b = 1.$$

Finally, we compute the required expression $$2a + b$$:

$$2a + b = 2(4) + 1 = 8 + 1 = 9.$$

Hence, the correct answer is Option B.

We begin by assuming that a common tangent to both curves can be written in the slope-intercept form $$y = mx + c$$, where $$m$$ is the slope and $$c$$ is the y-intercept. We will impose the condition of tangency separately on each curve and then combine the resulting relations for $$m$$ and $$c$$.

First, we impose tangency to the parabola $$y^{2}=16x$$. Substituting $$y = mx + c$$ in place of $$y$$ gives

$$\bigl(mx + c\bigr)^{2}=16x.$$

Expanding the left side we get

$$m^{2}x^{2}+2mc\,x+c^{2}-16x=0.$$

This is a quadratic equation in $$x$$. For a line to touch (and not cut) the parabola, this quadratic must have exactly one real root, i.e.\ its discriminant must be zero. The discriminant $$\Delta$$ of the quadratic $$A x^{2}+B x + C = 0$$ is $$\Delta=B^{2}-4AC$$. Here $$A=m^{2},\; B=2mc-16,\; C=c^{2}.$$ Setting $$\Delta=0$$, we obtain

$$(2mc - 16)^{2} - 4 m^{2} c^{2} = 0.$$

Expanding and simplifying term by term,

$$4m^{2}c^{2} - 64mc + 256 - 4m^{2}c^{2} = 0,$$

$$-64mc + 256 = 0,$$

$$mc = 4.$$

Next, we impose tangency to the hyperbola $$xy = -4$$. Substituting $$y = mx + c$$ gives

$$x(mx + c) = -4,$$

$$m x^{2} + c x + 4 = 0.$$

This too is a quadratic in $$x$$. Requiring tangency forces its discriminant to vanish:

$$c^{2} - 4 \cdot m \cdot 4 = 0,$$

$$c^{2} - 16m = 0,$$

$$c^{2} = 16m.$$

We now have a pair of simultaneous relations:

$$mc = 4 \quad \text{and} \quad c^{2} = 16m.$$

From $$mc = 4$$ we can express $$c$$ as $$c = \dfrac{4}{m}.$$ Substituting this into $$c^{2}=16m$$ gives

$$\left(\dfrac{4}{m}\right)^{2}=16m,$$

$$\dfrac{16}{m^{2}}=16m,$$

$$\dfrac{1}{m^{2}} = m,$$

$$m^{3}=1.$$

The real solution of $$m^{3}=1$$ is $$m=1$$ (the other two cubic roots are complex and therefore do not correspond to real tangents). Substituting $$m=1$$ back into $$c = \dfrac{4}{m}$$ we find

$$c = 4.$$

Thus the required common tangent is

$$y = 1\cdot x + 4,$$

or, after rearranging all terms to one side,

$$x - y + 4 = 0.$$

Because this equation matches Option B, we have located the single common tangent correctly.

Hence, the correct answer is Option B.

We have two subsets of the plane $$\mathbb R \times \mathbb R$$ defined as 

$$A=\{(a,b):|a-5|<1 \text{ and } |b-5|<1\}$$ and $$B=\{(a,b):4(a-6)^2+9(b-5)^2\le 36\}.$$

First we rewrite the description of each set in a form that is easier to interpret.

For set $$A$$, the inequalities $$|a-5|<1$$ and $$|b-5|<1$$ mean

$$4<a<6 \quad\text{and}\quad 4<b<6.$$

Thus $$A$$ is the open square whose centre is the point $$(5,5)$$ and whose side length is $$2$$.

For set $$B$$ we begin by recalling the standard form of an ellipse. The general formula

$$\frac{(x-h)^2}{p^2}+\frac{(y-k)^2}{q^2}\le 1$$

represents an ellipse centred at $$(h,k)$$ with semi-axes $$p$$ (along the $$x$$-direction) and $$q$$ (along the $$y$$-direction).

We rewrite the defining inequality of $$B$$ in this standard form. Starting from

$$4(a-6)^2+9(b-5)^2\le 36,$$

we divide every term by $$36$$:

$$\frac{4(a-6)^2}{36}+\frac{9(b-5)^2}{36}\le 1.$$

Simplifying the fractions, we obtain

$$\frac{(a-6)^2}{9}+\frac{(b-5)^2}{4}\le 1.$$

Hence $$B$$ is an ellipse with centre $$(6,5),$$ semi-major axis $$3$$ along the $$a$$-direction and semi-minor axis $$2$$ along the $$b$$-direction.

Now we test whether every point of $$A$$ also lies in $$B$$. So let us take an arbitrary point $$(a,b)\in A$$. By membership in $$A$$ we know

$$|a-5|<1 \; \Longrightarrow \; 4<a<6,$$

$$|b-5|<1 \; \Longrightarrow \; 4<b<6.$$

We next estimate the two squared distances that appear in the ellipse inequality.

Because $$4<a<6,$$ we have

$$|a-6|=6-a<6-4=2,$$

so

$$(a-6)^2<2^2=4.$$

Similarly, since $$4<b<6,$$ we get

$$|b-5|<1 \;\Longrightarrow\; (b-5)^2<1.$$

We substitute these upper bounds into the left-hand side of the ellipse condition:

$$\frac{(a-6)^2}{9}+\frac{(b-5)^2}{4}\;<\;\frac{4}{9}+\frac{1}{4}.$$

To compare this sum with $$1,$$ we find a common denominator $$36$$:

$$\frac{4}{9}+\frac{1}{4}=\frac{16}{36}+\frac{9}{36}=\frac{25}{36}.$$

Because $$\frac{25}{36}<1,$$ we have

$$\frac{(a-6)^2}{9}+\frac{(b-5)^2}{4}<1.$$

This strict inequality certainly implies the weak inequality

$$\frac{(a-6)^2}{9}+\frac{(b-5)^2}{4}\le 1,$$

which is exactly the condition for $$(a,b)$$ to belong to $$B$$. Therefore every point of $$A$$ lies inside the ellipse $$B$$; symbolically, $$A\subset B.$$

It remains to see that the reverse inclusion does not hold. Consider, for instance, the centre of the ellipse, the point $$P=(6,5).$$ Substituting into the ellipse equation gives

$$4(6-6)^2+9(5-5)^2=0\le 36,$$

so $$P\in B.$$ However, for this point we have $$|a-5|=|6-5|=1,$$ which is not strictly less than $$1,$$ hence $$P\notin A.$$ Thus $$B\not\subset A.$$

Combining these results we conclude

$$A\subset B \quad\text{but}\quad B\not\subset A.$$

Hence, the correct answer is Option C.

Let the square have vertices $$O, A, B, C$$ in that order, with $$O$$ at the origin. The length of every side is given to be $$2$$.

The side $$OA$$ passes through the origin and makes an angle of $$30^{\circ}$$ with the positive $$x$$-axis. Hence the direction ratios of $$OA$$ are $$\bigl(\cos 30^{\circ},\; \sin 30^{\circ}\bigr).$$ Using the length $$2,$$ the exact coordinates of $$A$$ are obtained by simple trigonometry: $$\begin{aligned} x_A &= 2\cos 30^{\circ}=2\left(\frac{\sqrt3}{2}\right)=\sqrt3,\\[2pt] y_A &= 2\sin 30^{\circ}=2\left(\frac12\right)=1. \end{aligned}$$ Thus $$A(\sqrt3,\;1).$$

Write the vector $$\overrightarrow{OA}$$ in component form: $$\overrightarrow{OA}=(\sqrt3,\,1).$$ To obtain the adjacent side of the square we need a vector of the same length that is perpendicular to $$\overrightarrow{OA}.$$ Rotating a vector $$(x,\,y)$$ through $$90^{\circ}$$ counter-clockwise produces $$(-y,\,x).$$ Therefore $$\overrightarrow{w}=(-1,\;\sqrt3)$$ is perpendicular to $$\overrightarrow{OA}$$ and has the same magnitude because $$\bigl|(-1,\sqrt3)\bigr|=\sqrt{(-1)^2+(\sqrt3)^2}=\sqrt{1+3}=2.$$

If instead we had rotated clockwise, we would get $$(1,-\sqrt3),$$ whose $$y$$-component is negative; that would force two vertices below the $$x$$-axis, contradicting the condition that the whole square lies above it. Hence we retain $$\overrightarrow{w}=(-1,\;\sqrt3).$$

The remaining two vertices are obtained by adding this vector once to $$A$$ and once to $$O$$:

For $$B:$$ $$\begin{aligned} B &= A+\overrightarrow{w}\\ &=\bigl(\sqrt3,\,1\bigr)+\bigl(-1,\,\sqrt3\bigr)\\ &=\bigl(\sqrt3-1,\,1+\sqrt3\bigr). \end{aligned}$$

For $$C:$$ $$\begin{aligned} C &= O+\overrightarrow{w}\\ &=\bigl(0,\,0\bigr)+\bigl(-1,\,\sqrt3\bigr)\\ &=\bigl(-1,\,\sqrt3\bigr). \end{aligned}$$

Collecting all four vertices with their $$x$$-coordinates:

$$\begin{aligned} O &: (0,\,0) &\Rightarrow&\ x_O=0,\\ A &: (\sqrt3,\,1) &\Rightarrow&\ x_A=\sqrt3,\\ B &: (\sqrt3-1,\,1+\sqrt3) &\Rightarrow&\ x_B=\sqrt3-1,\\ C &: (-1,\,\sqrt3) &\Rightarrow&\ x_C=-1. \end{aligned}$$

Now add these four $$x$$-coordinates explicitly:

$$\begin{aligned} x_O+x_A+x_B+x_C&=0+\sqrt3+(\sqrt3-1)+(-1)\\ &=\sqrt3+\sqrt3-1-1\\ &=2\sqrt3-2. \end{aligned}$$

Thus the required sum of the $$x$$-coordinates is $$2\sqrt3-2.$

Hence, the correct answer is Option A.

$$L_1: x - \sqrt{3}y = \alpha \quad (y > 0)$$

$$L_2: x + \sqrt{3}y = \alpha \quad (y < 0)$$

For $$L_1$$: $$y = \frac{1}{\sqrt{3}}x - \frac{\alpha}{\sqrt{3}} \implies \tan\theta_1 = \frac{1}{\sqrt{3}} \implies \theta_1 = 30^\circ$$

For $$L_2$$: $$y = -\frac{1}{\sqrt{3}}x + \frac{\alpha}{\sqrt{3}} \implies \tan\theta_2 = -\frac{1}{\sqrt{3}} \implies \theta_2 = -30^\circ$$

The total angle between the two half-lines at vertex $$P$$ is: $$\angle APB = 30^\circ - (-30^\circ) = 60^\circ$$

Given that $$PA = PB = \alpha$$ and the vertical angle $$\angle APB = 60^\circ$$, $$\triangle PAB$$ must be an equilateral triangle. Therefore, the side length of the triangle is $$AB = \alpha$$.

Since $$Q$$ lies on the angle bisector (the x-axis) and $$\triangle PAB$$ is equilateral, $$PQ$$ is the median/altitude of $$\triangle PAB$$.

$$PQ = PA \cos(30^\circ) = \alpha \frac{\sqrt{3}}{2}$$

$$\alpha \frac{\sqrt{3}}{2} = \frac{9}{2} \implies \alpha = \frac{9}{\sqrt{3}} = 3\sqrt{3}$$

Circumradius, $$R = \frac{a}{2\sin(60^\circ)} = \frac{\alpha}{\sqrt{3}}$$

$$R = \frac{3\sqrt{3}}{\sqrt{3}} = 3$$

$$\frac{\alpha^2}{R} = \frac{(3\sqrt{3})^2}{3} = \frac{27}{3} = 9$$

For the parabola $$y=x^2$$ the slope of the tangent is $$\frac{dy}{dx}=2x$$

Given,

$$2x=4$$

$$x=2$$

Hence,

$$P=(2,4)$$

For the circle $$x^2+y^2=2$$ differentiating, $$2x+2y\frac{dy}{dx}=0$$

$$\frac{dy}{dx}=-\frac{x}{y}$$

Given,

$$-\frac{x}{y}=-1$$

$$x=y$$

Since $$Q$$ lies on $$x^2+y^2=2$$

we get $$2x^2=2$$

$$x=y=1$$

Hence,

$$Q=(1,1)$$

For the ellipse $$x^2+4y^2=8$$ differentiating, $$2x+8y\frac{dy}{dx}=0$$

$$\frac{dy}{dx}=-\frac{x}{4y}$$

Given,

$$-\frac{x}{4y}=-\frac12$$

$$x=2y$$

Substituting into $$x^2+4y^2=8$$ gives $$4y^2+4y^2=8$$

$$8y^2=8$$

$$y=1,\quad x=2$$

Hence, $$R=(2,1)$$

Now, $$PR=|4-1|=3$$ and $$QR=|2-1|=1$$

Since $$PR$$ is vertical and $$QR$$ is horizontal,

$$\angle PRQ=90^\circ$$

Therefore, the circumradius of triangle $$PQR$$ is half the hypotenuse.

Now,

$$PQ=\sqrt{(2-1)^2+(4-1)^2}$$

$$=\sqrt{10}$$

Hence,

$$\text{Radius}=\frac{PQ}{2}$$

$$=\frac{\sqrt{10}}{2}$$

Therefore,

$$\boxed{\frac{\sqrt{10}}{2}}$$

Evaluating entry (P):

Radius $$R$$ equals the perpendicular distance from center $$(1,2)$$ to $$3x + 4y - 1 = 0$$:

$$R = \frac{\vert{}3(1) + 4(2) - 1\vert{}}{\sqrt{3^2 + 4^2}} = \frac{10}{5} = 2$$

$$\text{Equation of circle: } (x-1)^2 + (y-2)^2 = 4$$

Testing point (3,2) from List-II: $$(3-1)^2 + (2-2)^2 = 4 \implies 4 = 4 \implies \text{(P)} \rightarrow \text{(3)}$$

Evaluating entry (Q):

Equation of tangent to $$y^2 = 8x$$ with slope $$m$$: $$y = mx + \frac{2}{m}$$

Since it touches $$x^2 + y^2 = 2$$: $$\frac{\vert{}2/m\vert{}}{\sqrt{1 + m^2}} = \sqrt{2} \implies \frac{4}{m^2(1+m^2)} = 2 \implies m^4 + m^2 - 2 = 0 \implies m = 1 \quad (m > 0)$$

$$\text{Tangent line: } y = x + 2$$

Testing point (7,9) from List-II: $$9 = 7 + 2 \implies \text{(Q)} \rightarrow \text{(2)}$$

Evaluating entry (R):

$$\frac{x^2}{16} + \frac{y^2}{12} = 1 \implies a=4, \ b=\sqrt{12}$$

$$e = \sqrt{1 - \frac{12}{16}} = \frac{1}{2} \implies ae = 2$$

$$M = \left(ae, \frac{b^2}{a}\right) = \left(2, \frac{12}{4}\right) = (2, 3)$$

Equation of normal at $$M(2,3)$$: $$\frac{a^2x}{x_1} - \frac{b^2y}{y_1} = a^2 - b^2 \implies \frac{16x}{2} - \frac{12y}{3} = 16 - 12 \implies 2x - y = 1$$

Testing point (1,1) from List-II: $$2(1) - 1 = 1 \implies \text{(R)} \rightarrow \text{(1)}$$

Evaluating entry (S):

$$ae = 5, \quad \frac{a}{e} = \frac{16}{5} \implies a^2 = 16 \implies b^2 = a^2(e^2 - 1) = (ae)^2 - a^2 = 25 - 16 = 9$$

$$\text{Equation of hyperbola: } \frac{x^2}{16} - \frac{y^2}{9} = 1$$

Testing point $$(8, 3\sqrt{3})$$ from List-II: $$\frac{64}{16} - \frac{27}{9} = 4 - 3 = 1 \implies \text{(S)} \rightarrow \text{(5)}$$

We are given the midpoints of the sides of triangle $$ABC$$ as $$M_1 = \left(\frac{5}{2}, 7\right)$$, $$M_2 = \left(\frac{5}{2}, 3\right)$$, and $$M_3 = (4, 5)$$. Using the property that each vertex equals the sum of two adjacent midpoints minus the opposite midpoint:

$$A = M_1 + M_3 - M_2 = \left(\frac{5}{2} + 4 - \frac{5}{2},\; 7 + 5 - 3\right) = (4, 9)$$

$$B = M_1 + M_2 - M_3 = \left(\frac{5}{2} + \frac{5}{2} - 4,\; 7 + 3 - 5\right) = (1, 5)$$

$$C = M_2 + M_3 - M_1 = \left(\frac{5}{2} + 4 - \frac{5}{2},\; 3 + 5 - 7\right) = (4, 1)$$

We verify: midpoint of $$AB = \left(\frac{4+1}{2}, \frac{9+5}{2}\right) = \left(\frac{5}{2}, 7\right) = M_1$$, midpoint of $$BC = \left(\frac{1+4}{2}, \frac{5+1}{2}\right) = \left(\frac{5}{2}, 3\right) = M_2$$, midpoint of $$AC = \left(\frac{4+4}{2}, \frac{9+1}{2}\right) = (4, 5) = M_3$$.

Now we find the side lengths:

$$a = BC = \sqrt{(4-1)^2 + (1-5)^2} = \sqrt{9 + 16} = 5$$

$$b = AC = \sqrt{(4-4)^2 + (1-9)^2} = \sqrt{64} = 8$$

$$c = AB = \sqrt{(1-4)^2 + (5-9)^2} = \sqrt{9 + 16} = 5$$

The incenter is given by $$(h, k) = \left(\frac{a \cdot x_A + b \cdot x_B + c \cdot x_C}{a + b + c},\; \frac{a \cdot y_A + b \cdot y_B + c \cdot y_C}{a + b + c}\right)$$:

$$h = \frac{5(4) + 8(1) + 5(4)}{5 + 8 + 5} = \frac{20 + 8 + 20}{18} = \frac{48}{18} = \frac{8}{3}$$

$$k = \frac{5(9) + 8(5) + 5(1)}{18} = \frac{45 + 40 + 5}{18} = \frac{90}{18} = 5$$

Therefore, $$3h + k = 3 \times \frac{8}{3} + 5 = 8 + 5 = 13$$.

Hence, the correct answer is Option 3.

The circle is $$x^{2}+y^{2}-8x=0$$, which can be written as $$(x-4)^{2}+y^{2}=16$$. Its centre is $$(4,0)$$ and radius is $$4$$.

The hyperbola is $$\frac{x^{2}}{9}-\frac{y^{2}}{4}=1$$, or equivalently $$4x^{2}-9y^{2}=36$$ $$-(1)$$.

For the points of intersection A and B we must satisfy both the circle and the hyperbola. From the circle, $$y^{2}=8x-x^{2}$$ $$-(2)$$. Substituting $$y^{2}$$ from $$(2)$$ into $$(1)$$ gives

$$4x^{2}-9(8x-x^{2})=36$$ $$\Rightarrow 4x^{2}-72x+9x^{2}=36$$ $$\Rightarrow 13x^{2}-72x-36=0$$ $$-(3)$$.

The discriminant of $$(3)$$ is $$\Delta = (-72)^{2}-4\cdot 13\cdot (-36)=5184+1872=7056=84^{2}$$, so the roots are

$$x=\frac{72\pm 84}{26} \;\Rightarrow\; x_{1}=6,\; x_{2}=-\frac{6}{13}$$.

Using $$(2)$$ for each root: For $$x=6$$, $$y^{2}=8\cdot 6-6^{2}=48-36=12 \;\Rightarrow\; y=\pm 2\sqrt{3}$$. For $$x=-\dfrac{6}{13}$$, $$y^{2}=-\dfrac{660}{169}\lt 0$$, which is impossible. Hence the real intersection points are $$A(6,2\sqrt{3}),\; B(6,-2\sqrt{3})$$.

Let the moving point be $$P(x,y)$$ on the line $$2x-3y+4=0$$ $$-(4)$$.

The centroid $$G(X,Y)$$ of $$\triangle PAB$$ is $$X=\frac{x+6+6}{3}=\frac{x+12}{3}, \qquad Y=\frac{y+2\sqrt{3}-2\sqrt{3}}{3}=\frac{y}{3}$$ $$-(5)$$.

Express $$x$$ and $$y$$ from $$(5)$$: $$x=3X-12, \qquad y=3Y$$ $$-(6)$$.

Substitute $$(6)$$ into the line $$(4)$$:

$$2(3X-12)-3(3Y)+4=0$$ $$\Rightarrow 6X-24-9Y+4=0$$ $$\Rightarrow 6X-9Y-20=0$$ $$\Rightarrow 6X-9Y=20$$.

Replacing $$(X,Y)$$ by the usual $$x,y$$ gives the locus of the centroid: $$6x-9y=20$$.

Thus the required line is $$6x-9y=20$$, which corresponds to Option C.

We have $$\triangle PQR$$ with $$P(5, 4)$$, $$Q(-2, 4)$$, $$R(a, b)$$, area = 35 sq. units, and orthocentre $$O\left(2, \dfrac{14}{5}\right)$$.

Since $$P$$ and $$Q$$ have the same y-coordinate (y = 4), the base $$PQ$$ is horizontal with length $$|5 - (-2)| = 7$$.

Area = $$\dfrac{1}{2} \times \text{base} \times \text{height} = \dfrac{1}{2} \times 7 \times |b - 4| = 35$$.

The orthocentre has y-coordinate $$\dfrac{14}{5}$$, which is less than 4. Since PQ is horizontal (slope = 0), the altitude from R to PQ is vertical: $$x = a$$. The orthocentre lies on this altitude, so $$a = 2$$.

Slope of PR: $$\dfrac{b - 4}{a - 5} = \dfrac{b - 4}{2 - 5} = \dfrac{b - 4}{-3}$$.

The altitude from Q(-2, 4) is perpendicular to PR, so its slope = $$\dfrac{3}{b - 4}$$.

Altitude from Q(-2, 4) with slope $$\dfrac{3}{b-4}$$:

This passes through the orthocentre $$(2, 14/5)$$:

This is consistent with our area calculation ($$b = -6$$ gives $$|b - 4| = 10$$).

So $$R = (2, -6)$$.

The answer is $$\boxed{3}$$, which corresponds to Option 4.

The ellipse is $$\frac{x^{2}}{9}+\frac{y^{2}}{4}=1$$, so a parametric point on it is given as $$P(3\cos\alpha,\;2\sin\alpha)$$, with $$\alpha\neq0$$.

Let $$Q(x_{2},y_{2})$$ lie on the circle
$$(x-7)^{2}+(y-7)^{2}=16$$  $$-(1)$$
and let $$R(x_{3},y_{3})$$ lie on the straight line
$$x+y=5$$  $$-(2)$$.

The centroid $$G$$ of $$\triangle PQR$$ is given to be
$$G\left(2+\cos\alpha,\;3+\frac23\sin\alpha\right).$$
Using the centroid formula $$G\Bigl(\dfrac{x_{1}+x_{2}+x_{3}}{3},\dfrac{y_{1}+y_{2}+y_{3}}{3}\Bigr)$$, equate the coordinates:

$$\frac{3\cos\alpha+x_{2}+x_{3}}{3}=2+\cos\alpha,$$
$$\frac{2\sin\alpha+y_{2}+y_{3}}{3}=3+\frac23\sin\alpha.$$

Solving each for $$x_{3}$$ and $$y_{3}$$:

$$x_{3}=3(2+\cos\alpha)-(3\cos\alpha+x_{2})=6-x_{2},$$
$$y_{3}=3\!\left(3+\frac23\sin\alpha\right)-(2\sin\alpha+y_{2})=9-y_{2}.$$

Notice that both $$x_{3}$$ and $$y_{3}$$ are independent of $$\alpha$$. Hence

$$R(6-x_{2},\;9-y_{2}).$$

Point $$R$$ must also satisfy the line equation $$(2)$$:

$$(6-x_{2})+(9-y_{2})=5\quad\Longrightarrow\quad x_{2}+y_{2}=10.$$

Therefore, point $$Q(x_{2},y_{2})$$ must satisfy the simultaneous system

$$\begin{cases}(x-7)^{2}+(y-7)^{2}=16,\\ x+y=10.\end{cases}$$

Put $$y=10-x$$ into the circle $$-(1)$$:

$$(x-7)^{2}+\bigl[(10-x)-7\bigr]^{2}=16,$$
$$(x-7)^{2}+(3-x)^{2}=16,$$
$$x^{2}-14x+49+x^{2}-6x+9=16,$$
$$2x^{2}-20x+58=16,$$
$$2x^{2}-20x+42=0,$$
$$x^{2}-10x+21=0,$$
$$(x-3)(x-7)=0.$$

Thus
$$x_{2}=3\;\; \text{or}\;\; x_{2}=7.$$
Correspondingly $$y_{2}=10-x_{2}$$ gives

$$Q_{1}(3,7),\qquad Q_{2}(7,3).$$

Substituting each into $$R(6-x_{2},9-y_{2})$$:

For $$Q_{1}(3,7):\; R_{1}(6-3,\;9-7)=(3,2).$$
For $$Q_{2}(7,3):\; R_{2}(6-7,\;9-3)=(-1,6).$$

Both $$R_{1}$$ and $$R_{2}$$ satisfy the line $$x+y=5$$ as required.

Hence the ordinates (y-coordinates) of the possible points $$R$$ are $$2$$ and $$6$$.
Sum of ordinates = $$2+6=8.$$

Option D which is: $$8$$

The rectangular hyperbola is $$xy = 12$$.
Let $$P(x_1 , y_1)$$ and $$Q(x_2 , y_2)$$ lie on it, so

$$x_1y_1 = 12, \qquad x_2y_2 = 12$$ $$-(1)$$

The midpoint of $$PQ$$ is given as $$\left(\tfrac{1}{2}, -\tfrac{1}{2}\right)$$, hence

$$\frac{x_1+x_2}{2} = \frac12 \Longrightarrow x_1 + x_2 = 1$$
$$\frac{y_1+y_2}{2} = -\frac12 \Longrightarrow y_1 + y_2 = -1$$ $$-(2)$$

The area of $$\triangle OPQ$$ is

$$\text{Area} = \frac12\,\left|x_1y_2 - x_2y_1\right|$$ $$-(3)$$

To obtain $$x_1y_2 - x_2y_1$$, first evaluate the sum $$x_1y_2 + x_2y_1$$. Multiply the two equalities in $$(2)$$:

$$(x_1+x_2)(y_1+y_2)=1\cdot(-1)=-1$$

Expanding the left side:

$$(x_1y_1 + x_1y_2 + x_2y_1 + x_2y_2)= -1$$

Using $$(1)$$, $$x_1y_1 + x_2y_2 = 12 + 12 = 24$$, hence

$$24 + (x_1y_2 + x_2y_1) = -1 \Longrightarrow x_1y_2 + x_2y_1 = -25$$ $$-(4)$$

Now introduce the convenient parametrisation of the hyperbola: take $$P( t,\; \tfrac{12}{t})$$ and $$Q( s,\; \tfrac{12}{s})$$.

From $$(2)$$:

$$t+s = 1$$
$$\frac{12}{t} + \frac{12}{s} = -1 \;\Longrightarrow\; 12\Bigl(\frac{s+t}{st}\Bigr) = -1$$
Since $$s+t = 1$$, this gives $$\displaystyle \frac{12}{st} = -1 \;\Longrightarrow\; st = -12$$ $$-(5)$$

Thus $$t$$ and $$s$$ are the roots of $$u^2 - (s+t)u + st = 0$$:

$$u^2 - u - 12 = 0 \;\Longrightarrow\; (u-4)(u+3)=0$$

Therefore $$\{t,s\} = \{4,\,-3\}$$.

Choose $$P(4,3),\quad Q(-3,-4)$$ (the other assignment only swaps the labels and does not change the area).

Evaluate the determinant in $$(3)$$:

$$x_1y_2 = 4\cdot(-4) = -16,\qquad x_2y_1 = (-3)\cdot3 = -9$$
$$x_1y_2 - x_2y_1 = -16 - (-9) = -7$$

Hence

$$\text{Area} = \frac12 \bigl| -7 \bigr| = \frac{7}{2}$$

Option C which is: $$\frac{7}{2}$$

Let the two concurrent lines be$$4x-3y=12\alpha \qquad -(1)$$and$$4\alpha x+3\alpha y=12 \qquad -(2)$$with $$\alpha\neq 0$$.

Coordinates of the intersection
From $$(1)$$ we get $$4x-3y=12\alpha$$.
Divide $$(2)$$ by $$\alpha$$ to obtain $$4x+3y=\dfrac{12}{\alpha}\qquad -(3)$$.

Add $$(1)$$ and $$(3)$$:
$$8x=12\alpha+\dfrac{12}{\alpha}\;\Longrightarrow\;x=\dfrac{3}{2}\left(\alpha+\dfrac{1}{\alpha}\right).$$

Subtract $$(1)$$ from $$(3)$$:
$$6y=\dfrac{12}{\alpha}-12\alpha\;\Longrightarrow\;y=2\left(\dfrac{1}{\alpha}-\alpha\right).$$

Eliminating $$\alpha$$ to get the locus $$S$$
Let $$t=\alpha+\dfrac{1}{\alpha}=\dfrac{2}{3}x$$ and$$s=\alpha-\dfrac{1}{\alpha}=-\dfrac{y}{2}.$$ For any real $$\alpha\neq 0$$ we have $$(\alpha+\tfrac{1}{\alpha})^{2}-(\alpha-\tfrac{1}{\alpha})^{2}=4.$$ Substituting $$t$$ and $$s$$, $$\left(\dfrac{2}{3}x\right)^{2}-\left(-\dfrac{y}{2}\right)^{2}=4.$$ This simplifies to $$\dfrac{4}{9}x^{2}-\dfrac{y^{2}}{4}=4 \;\Longrightarrow\; \dfrac{x^{2}}{9}-\dfrac{y^{2}}{16}=1.$$ Hence $$S$$ is the hyperbola $$\boxed{\dfrac{x^{2}}{9}-\dfrac{y^{2}}{16}=1}.$

Equation of the required tangent $$T$$
The tangent is parallel to the line $$4x-\dfrac{3}{$$\sqrt$$2}y=0$$ whose slope is $$m=\dfrac{4$$\sqrt$$2}{3}.$$
Let the tangent cut the axes at $$(p,0)$$ and $$(0,q)\;(q\gt 0).$$ In intercept form it is $$\dfrac{x}{p}+\dfrac{y}{q}=1,$$ whose slope is $$-\dfrac{q}{p}.$$ Parallelism gives $$-\dfrac{q}{p}=m=\dfrac{4$$\sqrt$$2}{3}\;\Longrightarrow\;q=-mp.$$

Tangency condition
For the hyperbola $$\dfrac{x^{2}}{a^{2}}-\dfrac{y^{2}}{b^{2}}=1$$ with $$a^{2}=9,\;b^{2}=16,$$ the line $$y=mx+c$$ is a tangent iff $$c^{2}=a^{2}m^{2}-b^{2}.$$ Here $$y=mx+c$$ with $$c=-mp,$$ so $$c^{2}=(-mp)^{2}=m^{2}p^{2}.$$ Using the tangency criterion, $$m^{2}p^{2}=a^{2}m^{2}-b^{2} \;\Longrightarrow\; p^{2}=a^{2}-\dfrac{b^{2}}{m^{2}}.$$

Compute each term:
$$a^{2}m^{2}=9$$\left$$(\dfrac{4$$\sqrt$$2}{3}$$\right$$)^{2}=9$$\cdot$$\dfrac{32}{9}=32,$$ and therefore $$c^{2}=a^{2}m^{2}-b^{2}=32-16=16\quad$$\Rightarrow$$\quad c=4 \;(c\gt 0).$$

Since $$c=-mp,$$ we have $$4=-mp\;\Longrightarrow\;p=-\dfrac{4}{m} =-\dfrac{4}{\dfrac{4$$\sqrt$$2}{3}}=-\dfrac{3}{$$\sqrt$$2}.$$ Also $$q=-mp=-$$\left$$(\dfrac{4$$\sqrt$$2}{3}$$\right$$)\!$$\left$$(-\dfrac{3}{$$\sqrt$$2}$$\right$$)=4.$$

Required product
$$pq=$$\left$$(-\dfrac{3}{$$\sqrt$$2}$$\right$$)(4)=-\dfrac{12}{$$\sqrt$$2}=-6$$\sqrt$$2.$$

Thus the correct value is $$\boxed{-6$$\sqrt$$2}$$, which corresponds to
Option A.

The given data are:
A $$\left(1,2\right)$$, mid-point of $$AB$$ is $$\left(5,-1\right)$$, centroid $$G\left(3,4\right)$$.

1. Find the co-ordinates of $$B$$.
If $$B\left(x_B,y_B\right)$$, then using the mid-point formula
$$\frac{1+x_B}{2}=5,\;\; \frac{2+y_B}{2}=-1$$
$$\Rightarrow 1+x_B=10\;\; \text{and}\;\; 2+y_B=-2$$
$$\Rightarrow x_B=9,\;\; y_B=-4$$
Hence $$B\left(9,-4\right)$$.

2. Find the co-ordinates of $$C$$ from the centroid.
For $$C\left(x_C,y_C\right)$$, the centroid condition is
$$\frac{1+9+x_C}{3}=3,\;\; \frac{2-4+y_C}{3}=4$$
$$\Rightarrow 10+x_C=9\;\; \text{and}\;\; -2+y_C=12$$
$$\Rightarrow x_C=-1,\;\; y_C=14$$
Thus $$C\left(-1,14\right)$$.

3. Equation of the perpendicular bisector of $$AB$$.
Slope of $$AB$$: $$m_{AB}=\frac{-4-2}{9-1}=-\frac34$$
Perpendicular slope: $$\frac43$$.
Mid-point of $$AB$$ is $$\left(5,-1\right)$$, so
$$y+1=\frac43\bigl(x-5\bigr)\;\;-(1)$$

4. Equation of the perpendicular bisector of $$AC$$.
Slope of $$AC$$: $$m_{AC}=\frac{14-2}{-1-1}=-6$$
Perpendicular slope: $$\frac16$$.
Mid-point of $$AC$$ is $$\left(0,8\right)$$, so
$$y-8=\frac16\,x\;\;-(2)$$
or $$y=\frac16\,x+8$$.

5. Intersection of the two bisectors gives the circumcenter $$(\alpha,\beta)$$.
From $$(1)$$: $$y=\frac43\,(x-5)-1$$.
Set equal to $$(2)$$:
$$\frac43\,(x-5)-1=\frac16\,x+8$$
Multiply by $$6$$: $$8(x-5)-6=x+48$$
$$8x-40-6=x+48$$
$$8x-46=x+48$$
$$7x=94\;\;\Rightarrow\;\; x=\frac{94}{7}$$
Substitute in $$(2)$$:
$$y=\frac16\left(\frac{94}{7}\right)+8=\frac{94}{42}+8=\frac{47}{21}+8=\frac{215}{21}$$
Hence $$\alpha=\frac{94}{7},\;\; \beta=\frac{215}{21}$$.

6. Required value.
$$\alpha+\beta=\frac{94}{7}+\frac{215}{21}=\frac{282+215}{21}=\frac{497}{21}$$
Therefore $$21(\alpha+\beta)=21\cdot\frac{497}{21}=497$$.

Thus the correct choice is
Option C which is: $$497$$.

To find the maximum distance of point $$P$$ (on the circle) from the vertex of the parabola, we need to find the coordinates of the circle's center, its radius, and the vertex of the parabola.

The equation of the circle is $$x^2 + y^2 - 6x - 8y + 21 = 0$$.

We complete the square to find the center $$(h, k)$$ and radius $$r$$:

$$(x^2 - 6x + 9) + (y^2 - 8y + 16) = -21 + 9 + 16$$

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

• Center ($$C$$): $$(3, 4)$$
• Radius ($$r$$): $$\sqrt{4} = 2$$
• Vertex ($$V$$): $$(-3, -4)$$

The equation of the parabola is $$x^2 + 6x + y + 13 = 0$$.

Rearrange to find the vertex:

$$y = -x^2 - 6x - 13$$

$$y = -(x^2 + 6x + 9) - 13 + 9$$

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

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

For any point $$P$$ on a circle, the distance to an external point $$V$$ is maximized when $$P$$ lies on the line passing through the center $$C$$ and the point $$V$$, specifically on the "far side" of the circle.

The maximum distance $$d_{max}$$ is given by:

$$d_{max} = CV + r$$

First, calculate the distance between the center $$C(3, 4)$$ and the vertex $$V(-3, -4)$$ using the distance formula:

$$CV = \sqrt{(3 - (-3))^2 + (4 - (-4))^2}$$

$$CV = \sqrt{6^2 + 8^2}$$

$$CV = \sqrt{36 + 64}$$

$$CV = \sqrt{100} = 10$$

Now, add the radius of the circle:

$$d_{max} = 10 + 2 = 12$$

Conclusion:

The maximum distance of $$P$$ from the vertex of the parabola is 12.

Correct Option: C