Centre of a Circle from General Equation

1. Identify the properties of the circle
The given equation of the circle is:
$$x^2+y^2-6x-8y+21=0$$

We can rewrite this in standard form by completing the square for both $$x$$ and $$y$$:
$$(x^2-6x+9) + (y^2-8y+16) = -21 + 9 + 16$$
$$(x-3)^2 + (y-4)^2 = 4$$

From this equation, we determine:
Center of the circle $$C = (3,4)$$
Radius of the circle $$R = \sqrt{4} = 2$$

2. Utilize the condition for mutually perpendicular tangents
When a pair of tangents drawn from an external point $$P(h,k)$$ to a circle are mutually perpendicular, the point $$P$$ must lie on the director circle. The director circle is concentric with the original circle and has a radius equal to $$\sqrt{2}R$$.

Radius of the director circle $$R_d = \sqrt{2} \cdot 2 = 2\sqrt{2}$$

Therefore, the equation of the director circle is:
$$(x-3)^2 + (y-4)^2 = (2\sqrt{2})^2$$
$$(x-3)^2 + (y-4)^2 = 8$$

Since $$P(h,k)$$ lies on this director circle, it satisfies:
$$(h-3)^2 + (k-4)^2 = 8$$

3. Analyze the line constraint
We are given that $$P(h,k)$$ lies on the line:
$$4x-3y=0 \implies 4h-3k=0 \implies k = \frac{4}{3}h$$

Notice that if we substitute the coordinates of the center $$C(3,4)$$ into the line equation:
$$4(3) - 3(4) = 12 - 12 = 0$$

This means the line $$4x-3y=0$$ passes directly through the center of the circle $$C(3,4)$$. Consequently, the distance from the center $$C(3,4)$$ to the point $$P(h,k)$$ is exactly equal to the radius of the director circle $$R_d = 2\sqrt{2}$$.

4. Determine the coordinates of P
Substitute $$k = \frac{4}{3}h$$ into the director circle relation:
$$(h-3)^2 + \left(\frac{4}{3}h-4\right)^2 = 8$$
$$(h-3)^2 + \left[\frac{4}{3}(h-3)\right]^2 = 8$$
$$(h-3)^2 \left(1 + \frac{16}{9}\right) = 8$$
$$(h-3)^2 \left(\frac{25}{9}\right) = 8$$
$$(h-3)^2 = \frac{72}{25}$$

Taking the square root on both sides:
$$h-3 = \pm\frac{6\sqrt{2}}{5} \implies h = 3 \pm \frac{6\sqrt{2}}{5}$$

Since $$k = \frac{4}{3}h$$, the corresponding values for $$k$$ are:
$$k = \frac{4}{3}\left(3 \pm \frac{6\sqrt{2}}{5}\right) = 4 \pm \frac{8\sqrt{2}}{5}$$

5. Calculate the value of h^2+k^2
We expand the expression for $$h^2+k^2$$ using the two cases:

Case 1: Using the positive sign
$$h = 3 + \frac{6\sqrt{2}}{5}$$
$$k = 4 + \frac{8\sqrt{2}}{5}$$

$$h^2+k^2 = \left(3 + \frac{6\sqrt{2}}{5}\right)^2 + \left(4 + \frac{8\sqrt{2}}{5}\right)^2$$
$$h^2+k^2 = \left(9 + \frac{72}{25} + \frac{36\sqrt{2}}{5}\right) + \left(16 + \frac{128}{25} + \frac{64\sqrt{2}}{5}\right)$$
$$h^2+k^2 = 25 + \frac{200}{25} + \frac{100\sqrt{2}}{5}$$
$$h^2+k^2 = 25 + 8 + 20\sqrt{2} = 33 + 20\sqrt{2}$$

Case 2: Using the negative sign
$$h = 3 - \frac{6\sqrt{2}}{5}$$
$$k = 4 - \frac{8\sqrt{2}}{5}$$

$$h^2+k^2 = \left(3 - \frac{6\sqrt{2}}{5}\right)^2 + \left(4 - \frac{8\sqrt{2}}{5}\right)^2$$
$$h^2+k^2 = \left(9 + \frac{72}{25} - \frac{36\sqrt{2}}{5}\right) + \left(16 + \frac{128}{25} - \frac{64\sqrt{2}}{5}\right)$$
$$h^2+k^2 = 25 + \frac{200}{25} - \frac{100\sqrt{2}}{5}$$
$$h^2+k^2 = 25 + 8 - 20\sqrt{2} = 33 - 20\sqrt{2}$$

The general equation of a circle is $$x^{2}+y^{2}+2gx+2fy+c=0$$ with

• centre $$(-g,\,-f)$$ • radius $$R=\sqrt{g^{2}+f^{2}-c}$$

Because the circle passes through the origin $$(0,0)$$, substituting $$x=0,\,y=0$$ gives $$c=0$$. Hence every valid option must have the constant term $$0$$, which is true for all four choices.

For each option, find the centre, radius and the perpendicular distance to the line $$y=8$$. The circle will be tangent to the line only when

$$\text{distance from centre to line } y=8 = \text{radius}$$

Case A:

Equation $$x^{2}+y^{2}-6x-6y=0$$ gives $$2g=-6\Rightarrow g=-3$$, $$2f=-6\Rightarrow f=-3$$.
Centre $$(3,3)$$, radius $$R=\sqrt{(-3)^{2}+(-3)^{2}}=\sqrt{18}=3\sqrt{2}$$.
Distance to $$y=8$$ is $$|8-3|=5$$. Since $$5\neq 3\sqrt{2}$$, not tangent.

Case B:

Equation $$x^{2}+y^{2}-8x-6y=0$$ gives $$2g=-8\Rightarrow g=-4$$, $$2f=-6\Rightarrow f=-3$$.
Centre $$(4,3)$$, radius $$R=\sqrt{(-4)^{2}+(-3)^{2}}=\sqrt{25}=5$$.
Distance to $$y=8$$ is $$|8-3|=5$$, which equals the radius. Hence the circle is tangent to the line.

Case C:

Equation $$x^{2}+y^{2}-6x-8y=0$$ gives centre $$(3,4)$$ and radius $$R=\sqrt{(-3)^{2}+(-4)^{2}}=5$$.
Distance to $$y=8$$ is $$|8-4|=4$$. Here $$4\neq5$$, so not tangent.

Case D:

Equation $$x^{2}+y^{2}-10x-6y=0$$ gives centre $$(5,3)$$ and radius $$R=\sqrt{(-5)^{2}+(-3)^{2}}=\sqrt{34}$$.
Distance to $$y=8$$ is $$|8-3|=5$$. Since $$5\neq\sqrt{34}$$, not tangent.

Only Case B satisfies both conditions: passes through the origin and is tangent to the line $$y=8$$ with radius $$5$$.

Therefore, the required equation of the circle is:
Option B which is: $$x^{2}+y^{2}-8x-6y=0$$