We are given a fixed point A = $(3, 0)$ and a fixed circle $x^2 + y^2 = 36$ with center O = $(0, 0)$ and radius $R = 6$. Let the point B have coordinates $(x, y)$. The circle with diameter AB touches the fixed circle internally.
The center M of the variable circle (with diameter AB) is the midpoint of AB:
M = $\left(\frac{x+3}{2}, \frac{0+y}{2}\right) = \left(\frac{x+3}{2}, \frac{y}{2}\right)$
The radius $r$ of this variable circle is half the length of the diameter AB:
$r = \frac{1}{2} \text{length}(AB) = \frac{1}{2} \sqrt{(x-3)^2 + (y-0)^2} = \frac{1}{2} \sqrt{(x-3)^2 + y^2}$
For the variable circle to touch the fixed circle internally, the distance between their centers (OM) must equal the difference between their radii ($R - r$):
Distance OM = $\sqrt{\left(\frac{x+3}{2} - 0\right)^2 + \left(\frac{y}{2} - 0\right)^2} = R - r$
Substituting the expressions for OM, R, and r:
$\sqrt{\frac{(x+3)^2}{4} + \frac{y^2}{4}} = 6 - \frac{1}{2} \sqrt{(x-3)^2 + y^2}$
Simplify the equation:
$\frac{1}{2} \sqrt{(x+3)^2 + y^2} = 6 - \frac{1}{2} \sqrt{(x-3)^2 + y^2}$
Multiply both sides by 2:
$\sqrt{(x+3)^2 + y^2} = 12 - \sqrt{(x-3)^2 + y^2}$
Rearrange the terms:
$\sqrt{(x+3)^2 + y^2} + \sqrt{(x-3)^2 + y^2} = 12$
The equation $\sqrt{(x+3)^2 + y^2} + \sqrt{(x-3)^2 + y^2} = 12$ is the standard definition of an ellipse.
It represents the set of all points $(x, y)$ such that the sum of the distances from $(x, y)$ to two fixed points (the foci) is constant.
Here, the foci are $F_1 = (3, 0)$ and $F_2 = (-3, 0)$.
From the standard form of an ellipse with foci at $(\pm c, 0)$ and the sum of distances $2a$:
The eccentricity $e$ of an ellipse is defined as $e = \frac{c}{a}$.
Using the values calculated:
$e = \frac{3}{6} = \frac{1}{2}$
The question asks for the value of $72e^2$.
$72e^2 = 72 \times \left(\frac{1}{2}\right)^2$
$72e^2 = 72 \times \frac{1}{4}$
$72e^2 = 18$