Eccentricity of Locus Curve of Variable Diametric Circles

To find the eccentricity $e$ of the locus curve $C$ of the point $B$, we proceed as follows:

Let $A$ be the fixed point $(3, 0)$ and let $B(h, k)$ be a variable point whose locus we need to find.

1. Center and Radius of the Variable Circle: The circle has $AB$ as its diameter.

   • The center $M$ of this circle is the midpoint of $AB$: $M = \left( \frac{h+3}{2}, \frac{k}{2} \right)$
   • The radius $r$ of this circle is half the distance $AB$: $r = \frac{1}{2} AB = \frac{1}{2} \sqrt{(h-3)^2 + k^2}$

2. Condition for Internal Tangency: The given circle is $x^2 + y^2 = 36$, which has its center at the origin $O(0, 0)$ and a radius $R = 6$. Since the variable circle touches the circle $x^2 + y^2 = 36$ internally, the distance between their centers $OM$ must equal the difference of their radii: $OM = R - r \implies OM + r = R = 6$

3. Deriving the Locus Equation: The distance $OM$ from $O(0,0)$ to $M\left(\frac{h+3}{2}, \frac{k}{2}\right)$ is: $OM = \sqrt{\left( \frac{h+3}{2} \right)^2 + \left( \frac{k}{2} \right)^2} = \frac{1}{2} \sqrt{(h+3)^2 + k^2}$

   Substituting $OM$ and $r$ into the condition $OM + r = 6$: $\frac{1}{2} \sqrt{(h+3)^2 + k^2} + \frac{1}{2} \sqrt{(h-3)^2 + k^2} = 6$

   Multiplying both sides by $2$: $\sqrt{(h+3)^2 + k^2} + \sqrt{(h-3)^2 + k^2} = 12$

   Replacing $(h, k)$ with $(x, y)$, the equation of the locus curve $C$ is: $\sqrt{(x+3)^2 + y^2} + \sqrt{(x-3)^2 + y^2} = 12$

4. Calculating the Eccentricity: The equation represents an ellipse where the sum of the distances from any point $(x, y)$ on the curve to two fixed points $S_1(-3, 0)$ and $S_2(3, 0)$ is constant and equal to $12$.

   • Major axis length $2a = 12 \implies a = 6$
   • Distance between foci $2ae = 3 - (-3) = 6 \implies ae = 3$

   Therefore, the eccentricity $e$ is: $e = \frac{ae}{a} = \frac{3}{6} = \frac{1}{2}$

5. Final Value: We are required to compute $72e^2$: $72e^2 = 72 \times \left( \frac{1}{2} \right)^2 = 72 \times \frac{1}{4} = 18$