Go Round and Round

Let the above diagram represent one of the positions of the black circle. Let the radius (at the instant) of the black circle be $BD = r$ , where $B$ is the centre of the black circle. Let $A$ and $C$ denote the centres of the red and blue circles respectively. From the diagram, we see that

$AB = AE + EB = 3+r \\ BC = CD - BD = 10-r$

We find that $AB+BC = 13$ is a constant. Hence the locus of $B$ is an ellipse with major axis length $2a = 13$ , i.e., length of semi-major axis $a = \dfrac{13}{2}$ .

The points $A$ and $C$ are the focii of the ellipse and the distance between them is $2ae = 5$ , where $e$ denotes the eccentricity of the ellipse. From the values given above we get $e = \dfrac{5}{13}$ .

Using the identity $b^2 = a^2 (1-e^2)$ , we get $b = 6$ , where $b$ is the length of semi-minor axis of the ellipse.

The area of the ellipse gives the required area of the closed curve given by $A = \pi ab = \pi \cdot \dfrac{13}{2} \cdot 6 = 39 \pi$ . The answer is thus $\boxed{39}$ .

Applying the Cosine Rule, the radius of the third circle in the diagram is $r$ , where $(r+3)^2 \; = \; (10 - r)^2 + 5^2 - 10(10-r)\cos\theta$ and hence $r \; = \; \frac{58 - 50\cos\theta}{13 - 5\cos\theta}$

Thus the centre of the third circle has polar coordinates $\left( 10 - \frac{58 - 50\cos\theta}{13-5\cos\theta} \,,\,\theta\right) \; = \; \left(\frac{72}{13 - 5\cos\theta},\theta\right)$ Thus the desired area is $A \; = \; \tfrac12\int_0^{2\pi} \left(\frac{72}{13 - 5\cos\theta}\right)^2\,d\theta \; = \; 2592\int_0^{2\pi} \frac{d\theta}{(13 - 5\cos\theta)^2}$ Making the complex substitution $z = e^{i\theta}$ , we obtain $A \; = \; \frac{10368}{i}\int_{|z|=1} \frac{z\,dz}{(5z-1)^2(z-5)^2} \; = \; 20736\pi \mathrm{Res}_{z=\frac15} \frac{z}{(5z-1)^2(z-5)^2}$ and hence $A \; = \; \frac{20736\pi}{25} \left(\frac{d}{dz} \frac{z}{(z-5)^2}\right){\Huge|}_{z=\frac15} \; = \; -\frac{20736\pi(z+5)}{25(z-5)^3}{\Huge|}_{z=\frac15} \; = \; \frac{20736\pi}{25} \times \frac{26}{5} \times \frac{125}{24^3} \; = \; 39\pi$ making the answer $\boxed{39}$ .

Just for fun: