10.137. A chord $AB$ of constant length slides with its ends along a circle of radius $R$. A point $C$ on this chord, located at distances $a$ and $b$ from the ends $A$ and $B$ of the chord, describes a circle when the chord completes a full revolution. Calculate the area of the annulus enclosed between the given circle and the circle described by point $C$.
## Solution.

$\triangle A O B$ is isosceles (Fig. 10.123). Therefore, $\angle B A O = \angle A B O = \alpha$. Let $O C = r$. By the cosine theorem, $r^{2} = R^{2} + a^{2} - 2 a R \cos \alpha$. On the other hand, $r^{2} = R^{2} + b^{2} - 2 b R \cos \alpha$. Equating these, we get $a^{2} - b^{2} = 2 R \cos \alpha (a - b)$, from which $\cos \alpha = \frac{a + b}{2 R}$. Therefore, $r^{2} = R^{2} + b^{2} - 2 b R \frac{a + b}{2 R} = R^{2} - a b$. The areas of the circles are $S_{1} = \pi R^{2}, S_{2} = \pi r^{2} = \pi (R^{2} - a b)$. The area of the ring is $S = S_{1} - S_{2} = \pi R^{2} - \pi R^{2} + \pi a b = \pi a b$.

Answer: $\pi a b$.