[Trigonometric ratios in a right triangle] [Inscribed and circumscribed circles $]$

Point $O$ lies on segment $A B$, with $A O=13, O B=7$. A circle with center at point $O$ and radius 5 is drawn. Tangents are drawn from $A$ and $B$ to this circle, intersecting at point $M$, and the points of tangency lie on the same side of line $A B$. Find the radius of the circle circumscribed around triangle $A M B$.
Apply the formula $a=2 R \sin \alpha$.

## Solution

Let $C$ and $D$ be the points of tangency of the given circle with the sides $A M$ and $B M$ of triangle $A M B$, respectively.

Let $\angle M A B=\alpha, \angle M B A=\beta$.

From the right triangles $A C O$ and $D B O$, we find that

\[
\begin{aligned}
& \sin \alpha=\frac{C O}{A O}=\frac{5}{13}, \cos \alpha=\frac{12}{13} \\
& \sin \beta=\frac{O D}{O B}=\frac{5}{7}, \cos \beta=\frac{2 \sqrt{6}}{7}
\end{aligned}
\]

Then

\[
\begin{aligned}
& \sin \angle A M B=\sin \left(180^{\circ}-\alpha-\beta\right)=\sin (\alpha+\beta)= \\
& =\sin \alpha \cos \beta+\sin \beta \cos \alpha=\frac{10(6+\sqrt{6})}{91}
\end{aligned}
\]

Therefore, the desired radius is

\[
\frac{A B}{2 \sin \angle A M B}=\frac{20}{2 \cdot 10 \cdot \frac{6+\sqrt{6}}{91}}=\frac{91(6-\sqrt{6})}{30}
\]

![](https://cdn.mathpix.com/cropped/2024_05_06_0e468145bb05e34a17aag-38.jpg?height=406&width=838&top_left_y=1778&top_left_x=610)

## Answer

$\frac{91(6-\sqrt{6})}{30}$.