Let A be the point (3, 0) and circles with va | vedclass

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(B) Let $B = (x, y)$. The diameter of the circle is $AB$, where $A = (3, 0)$.The center of the circle is $M = (\frac{x+3}{2}, \frac{y}{2})$ and its ra

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$32$

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(B) Let $B = (x, y)$. The diameter of the circle is $AB$, where $A = (3, 0)$.The center of the circle is $M = (\frac{x+3}{2}, \frac{y}{2})$ and its radius is $r = \frac{1}{2} \sqrt{(x-3)^2 + y^2}$.The circle $x^2 + y^2 = 36$ has center $O = (0, 0)$ and radius $R = 6$.Since the circles touch internally, the distance between centers $OM = R - r$.$OM = \sqrt{(\frac{x+3}{2})^2 + (\frac{y}{2})^2} = 6 - \frac{1}{2} \sqrt{(x-3)^2 + y^2}$.Multiplying by $2$: $\sqrt{(x+3)^2 + y^2} = 12 - \sqrt{(x-3)^2 + y^2}$.Let $d_1 = \sqrt{(x-3)^2 + y^2}$ (distance $AB$) and $d_2 = \sqrt{(x+3)^2 + y^2}$ (distance $OB$).The equation is $d_2 = 12 - d_1$, or $d_1 + d_2 = 12$.Since $d_1 + d_2 = 12 > AB = 6$, the locus of $B$ is an ellipse with foci at $A(3, 0)$ and $O(-3, 0)$.The distance between foci $2ae = 6$, so $ae = 3$.The major axis $2a = 12$, so $a = 6$.Thus, $e = \frac{3}{6} = \frac{1}{2}$.Then $e^2 = \frac{1}{4}$.Finally, $72e^2 = 72 	imes \frac{1}{4} = 18$.

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 . . . . . .

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