JEE Main 2026 — Ellipse Question with Solution

Let the coordinates of point B be $(h,k)$.

The center of the circle with diameter AB is $C_1\left(\frac{h+3}{2},\frac{k}{2}\right)$ and its radius is $r_1=\frac{1}{2}\sqrt{(h-3{)}^2+k^2}$.

The given circle is $x^2+y^2=36$, which has center $C_2(0,0)$ and radius $r_2=6$.

Since the circles touch internally, the distance between their centers is equal to the difference of their radii:

$C_1{C}_2=r_2-r_1$

$\sqrt{{\left(\frac{h+3}{2}\right)}^2+{\left(\frac{k}{2}\right)}^2}=6-\frac{1}{2}\sqrt{(h-3{)}^2+k^2}$

Multiplying the entire equation by 2, we get:

$\sqrt{(h+3{)}^2+k^2}+\sqrt{(h-3{)}^2+k^2}=12$

This equation represents the locus of a point $B(h,k)$ such that the sum of its distances from two fixed points $S_1(-3,0)$ and $S_2(3,0)$ is constant and equal to $12$. This is the standard definition of an ellipse.

Thus, the foci of the ellipse are $(\pm 3,0)$ and the length of the major axis is $2a=12$.

The distance between the foci is $2ae=6$.

Substituting $2a=12$, we get $12e=6\Rightarrow e=\frac{1}{2}$.

We need to find the value of $72e^2$:

$72e^2=72{\left(\frac{1}{2}\right)}^2=72\times \frac{1}{4}=18$

Answer: $18$