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 $72e^2$ is equal to _______.

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

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

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:

$C_1C_2 = r_2 - r_1$

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

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 $12e = 6 \Rightarrow e = \dfrac{1}{2}$.

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

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

Answer: $18$