A circle of radius $r$ is inscribed in an angle equal to $\alpha$. Another circle of radius $R$ touches one side of the angle at the same point as the first, intersecting the other side of the angle at points $A$ and $B$. Find $AB$.
Let $O$ and $O_{1}$ be the centers of circles with radii $r$ and $R$, $K$ be the point of tangency of the first circle and the line $A B$, $P$ be the common point of tangency, $B M P$ be the given angle, and $F$ be the midpoint of $A B$. Then $A B^{2}=4\left(O_{1} B^{2}-O_{1} F^{2}\right)=4\left(R^{2}-(r-(R-r) \cos \right.$ $\left.\alpha)^{2}\right)=16 \cos ^{2} \alpha / 2\left((R-r)\left(R \sin ^{2} \alpha / 2+r \cos ^{2} \alpha / 2\right)\right.$.

![](https://cdn.mathpix.com/cropped/2024_05_06_9e7ee8b57660b9721937g-41.jpg?height=426&width=723&top_left_y=2065&top_left_x=679)

## Answer

$4 \cos \frac{\alpha}{2} \sqrt{(R-r)\left(R \sin ^{2} \frac{\alpha}{2}+r \cos ^{2} \frac{\alpha}{2}\right)}$.