A circle passes through vertices $A$ and $B$ of triangle $A B C$ and is tangent to line $A C$ at point $A$. Find the radius of the circle if $\angle B A C=\alpha, \angle A B C=\beta$ and the area of triangle $A B C$ is $S$.
Let $D$ be the point of intersection of the given circle with side $B C$. Knowing the area of triangle $A B C$, find $A B$ using the Law of Sines. Then prove that $\angle A D B=180^{\circ}-\alpha$.

## Solution

Let $D$ be the point of intersection of the given circle with line $B C$. Denote $A B=c, B C=a$. Applying the Law of Sines to triangle $A B C$, we get the proportion

$$
\frac{a}{\sin \alpha}=\frac{c}{\sin \left(180^{\circ}-\alpha-\beta\right)}=\frac{c}{\sin (\alpha+\beta)}
$$

from which $a=\frac{c \sin \alpha}{\sin (\alpha+\beta)}$. Then

$$
S=S_{\Delta \mathrm{ABC}}=\frac{1}{2} a c \sin \beta=\frac{c^{2} \sin \alpha \sin \beta}{2 \sin (\alpha+\beta)}
$$

from which we find that $c=\sqrt{\frac{2 S \sin (\alpha+\beta)}{\sin \alpha \sin \beta}}$. By the theorem on the angle between a tangent and a chord, we find that either $\angle A D B=\angle B A C=\alpha$ (Fig.1), or $\angle A D B=180^{\circ}-\angle B A C=180^{\circ}-\alpha$ (Fig.2). In both cases, $\sin \angle A D B=\sin \boldsymbol{\alpha}$. Let $R$ be the radius of the circle circumscribed around triangle $A B D$. Then

$$
R=\frac{A B}{2 \sin \angle A D B}=\frac{c}{2 \sin \alpha}=\sqrt{\frac{S \sin (\alpha+\beta)}{2 \sin ^{3} \alpha \sin \beta}}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_aa2960437c2efca8c53fg-20.jpg?height=435&width=598&top_left_y=640&top_left_x=730)

## Answer

$\sqrt{\frac{S \sin (\alpha+\beta)}{2 \sin ^{3} \alpha \sin \beta}}$.