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 fixed circle be $$x^{2}+y^{2}=36$$, whose centre is $$O(0,0)$$ and radius is $$R=6$$.

Point $$A$$ is fixed at $$A(3,0)$$. Take any point $$B(x,y)$$ and draw the circle having $$AB$$ as diameter. For this variable circle:

Centre $$M$$ is the midpoint of $$AB$$:
$$M\left(\frac{3+x}{2},\;\frac{y}{2}\right)$$

Radius $$r$$ equals half the length of $$AB$$:
$$r=\frac{1}{2}\sqrt{(x-3)^{2}+y^{2}}$$

The condition “touches the circle $$x^{2}+y^{2}=36$$ internally’’ means

$$OM+r=R \quad -(1)$$

Compute each term in $$(1)$$.

Distance $$OM$$:
$$OM=\sqrt{\left(\frac{3+x}{2}\right)^{2}+\left(\frac{y}{2}\right)^{2}}=\frac{1}{2}\sqrt{(x+3)^{2}+y^{2}}$$

Substitute in $$(1)$$:

$$\frac{1}{2}\sqrt{(x+3)^{2}+y^{2}}+\frac{1}{2}\sqrt{(x-3)^{2}+y^{2}}=6$$

Multiply by $$2$$:

$$\sqrt{(x+3)^{2}+y^{2}}+\sqrt{(x-3)^{2}+y^{2}}=12 \quad -(2)$$

Equation $$(2)$$ is the definition of an ellipse: the sum of distances of a point $$B(x,y)$$ from two fixed points is constant. Here the fixed points are $$F_{1}(-3,0)$$ and $$F_{2}(3,0)$$. Thus the curve $$C$$ is an ellipse with:

Distance between the foci $$=2c=|F_{1}F_{2}|=6 \;\Longrightarrow\; c=3$$

Major axis length $$=2a=12 \;\Longrightarrow\; a=6$$

Eccentricity is $$e=\dfrac{c}{a}=\dfrac{3}{6}=\dfrac{1}{2}$$

Therefore,
$$72e^{2}=72\left(\frac{1}{2}\right)^{2}=72\left(\frac{1}{4}\right)=18$$

Hence the required value is 18.