Let A be the point (3, 0) and circles with variable diameter AB...

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}$.<br><br>The given circle is$x^2 + y^2 = 36$, which has center$C_2(0, 0)$and radius$r_2 = 6$.<br><br>Since the circles touch internally, the distance between their centers is equal to the difference of their radii:<br><br>$C_1C_2 = r_2 - r_1$<br><br>$\sqrt{\left(\dfrac{h+3}{2}\right)^2 + \left(\dfrac{k}{2}\right)^2} = 6 - \dfrac{1}{2} \sqrt{(h-3)^2 + k^2}$<br><br>Multiplying the entire equation by 2, we get:<br><br>$\sqrt{(h+3)^2 + k^2} + \sqrt{(h-3)^2 + k^2} = 12$<br><br>This equation represents the locus of a point$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.<br><br>Thus, the foci of the ellipse are$(\pm 3, 0)$and the length of the major axis is$2a = 12$.<br><br>The distance between the foci is$2ae = 6$.<br><br>Substituting$2a = 12$, we get$12e = 6 \Rightarrow e = \dfrac{1}{2}$.<br><br>We need to find the value of$72e^2$:<br><br>$72e^2 = 72 \left(\dfrac{1}{2}\right)^2 = 72 \times \dfrac{1}{4} = 18$<br><br>Answer:$18