10.052. A circle touches one of the legs of an isosceles right triangle and passes through the vertex of the opposite acute angle. Find the radius of the circle if its center lies on the hypotenuse of the triangle, and the leg of the triangle is equal to $a$.
## Solution.

By the condition $C B=a, A C=C B$ (Fig. 10.52). Therefore, $\angle C A B=\angle C B A=45^{\circ}$ (since $\angle C=90^{\circ}$) $\Rightarrow A B=a \sqrt{2} \cdot \triangle A C B \sim \triangle O E B \Rightarrow \frac{a \sqrt{2}}{a \sqrt{2}-R}=\frac{a}{R}$, $R=\frac{a \sqrt{2}}{\sqrt{2}+1}=a(2-\sqrt{2})$

Answer: $\quad a(2-\sqrt{2})$