JEE Main 2026 — Apr 2, Shift 2 — Mathematics solution

JEE Main2026Apr 2, Shift 2Mathematics

Let $A$ be the point $(3, 0)$ and circles with variable diameter $A B$ touch the circle $x^{2} + y^{2} = 36$ internally. Let the curve $C$ be the locus of the point $B$ . If the eccentricity of $C$ is $e$ , then $72 e^{2}$ is equal to ____.

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Solution

Let $B = (h, k)$ . The variable circle has $A B$ as diameter, so its centre is the midpoint $M = (\frac{h + 3}{2}, \frac{k}{2})$ with radius $\frac{∣ A B ∣}{2}$ .

Internal tangency to the fixed circle (centre $O$ , radius $6$ ) requires $∣ O M ∣ = 6 - \frac{∣ A B ∣}{2}$ .

Squaring and simplifying ultimately gives $(h + 3)^{2} + k^{2} + (h - 3)^{2} + k^{2} = 12$ , which is the definition of an ellipse with foci $(\pm 3, 0)$ and sum of focal distances $2 a = 12$ .

So $a = 6$ , $c = 3$ , and eccentricity $e = \frac{c}{a} = \frac{1}{2}$ .

$∴ 72 e^{2} = 72 \cdot \frac{1}{4} = 18$ .

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