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…

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

The center of the circle with diameter AB is $C_1 \left( \dfrac{h+3}{2}, \dfrac{k}{2} \right)$C1​(2h+3​,2k​) and its radius is $r_1 = \dfrac{1}{2} \sqrt{(h-3)^2 + k^2}$r1​=21​(h−3)2+k2​.

The given circle is $x^2 + y^2 = 36$x2+y2=36, which has center $C_2(0, 0)$C2​(0,0) and radius $r_2 = 6$r2​=6.

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

$$C_1C_2 = r_2 - r_1$$C1​C2​=r2​−r1​

$$\sqrt{\left(\dfrac{h+3}{2}\right)^2 + \left(\dfrac{k}{2}\right)^2} = 6 - \dfrac{1}{2} \sqrt{(h-3)^2 + k^2}$$(2h+3​)2+(2k​)2​=6−21​(h−3)2+k2​

Multiplying the entire equation by 2, we get:

$$\sqrt{(h+3)^2 + k^2} + \sqrt{(h-3)^2 + k^2} = 12$$(h+3)2+k2​+(h−3)2+k2​=12

This equation represents the locus of a point $B(h, k)$B(h,k) such that the sum of its distances from two fixed points $S_1(-3, 0)$S1​(−3,0) and $S_2(3, 0)$S2​(3,0) is constant and equal to $12$12. This is the standard definition of an ellipse.

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

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

Substituting $2a = 12$2a=12, we get $12e = 6 \Rightarrow e = \dfrac{1}{2}$12e=6⇒e=21​.

We need to find the value of $72e^2$72e2:

$$72e^2 = 72 \left(\dfrac{1}{2}\right)^2 = 72 \times \dfrac{1}{4} = 18$$72e2=72(21​)2=72×41​=18

Answer: $18$18