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

JEE Main · 2026 · 02 Apr 2026 · Shift 2 · Ellipse

Let A be the point

(3, 0)

and circles with variable diameter AB 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 ______.

Answer: 18

Let

B = (h, k)

. The circle on

A B

as diameter has centre

(\frac{h + 3}{2}, \frac{k}{2})

and radius

\frac{1}{2} (h - 3)^{2} + k^{2}

. Internal contact with

x^{2} + y^{2} = 36

requires the distance from the origin to this centre to equal

6

minus the radius:

\frac{1}{2} (h + 3)^{2} + k^{2} = 6 - \frac{1}{2} (h - 3)^{2} + k^{2},

which rearranges to

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

. This is an ellipse with foci

(\pm 3, 0)

and

2 a = 12

, so

a = 6, c = 3

and

e = \frac{c}{a} = \frac{1}{2}

. Therefore

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

JEE Main · 2026 · 02 Apr 2026 · Shift 2 · Integer answer