JEE Mathematics 2023 Question with Solution

Given: The ellipse is

$$\frac{x^2}{36} + \frac{y^2}{9} = 1$$

and the circle is centered at $(2,0)$.

Find: The radius $r$ of the largest circle centered at $(2,0)$ and then the value of $12r^2$.

For the largest inscribed circle with a fixed center inside the ellipse, the radius is the perpendicular distance from the center to the ellipse along the normal at the nearest point.

Let a point on the ellipse be

$$P(6\cos\theta, 3\sin\theta)$$

The equation of the normal at $P$ is

$$(6\sec\theta)x - (3\csc\theta)y = 27$$

Since the normal passes through the center $(2,0)$, substitute this point:

$$(6\sec\theta)(2) - (3\csc\theta)(0) = 27$$

So,

$$12\sec\theta = 27$$

which gives

$$\sec\theta = \frac{9}{4}, \qquad \cos\theta = \frac{4}{9}$$

Now,

$$\sin\theta = \sqrt{1-\cos^2\theta} = \sqrt{1-\left(\frac{4}{9}\right)^2} = \frac{\sqrt{65}}{9}$$

Hence the point $P$ is

$$P = \left(6\cdot \frac{4}{9}, 3\cdot \frac{\sqrt{65}}{9}\right)=\left(\frac{8}{3}, \frac{\sqrt{65}}{3}\right)$$

Therefore the required radius is the distance from $(2,0)$ to $P$:

$$r = \sqrt{\left(\frac{8}{3}-2\right)^2 + \left(\frac{\sqrt{65}}{3}\right)^2}$$ $$r = \sqrt{\left(\frac{2}{3}\right)^2 + \frac{65}{9}} = \sqrt{\frac{69}{9}} = \frac{\sqrt{69}}{3}$$

Now compute

$$12r^2 = 12\left(\frac{\sqrt{69}}{3}\right)^2 = 12\cdot \frac{69}{9} = 92$$

Therefore, the value of $12r^2$ is $92$.

The solution working gives $92$, but the source the solution marks option A. Since the working is the primary source, the defensible answer is the option corresponding to $92$, which is D.