3. (6) Two unequal circles $w_{1}$ and $w_{2}$ with radii $r_{1}$ and $r_{2}$ respectively intersect at points $A$ and $B$. A point $O$ is taken on the plane such that $\angle O A B$ is $90^{\circ}$, and a circle $w$ with center $O$ is drawn, internally tangent to circles $w_{1}$ and $w_{2}$. Find the radius of circle $w$.
## 3. Answer: $r_{1}+r_{2}$.

It is easy to see that the circle $\omega$ can only touch $\omega_{1}, \omega_{2}$ externally. Without loss of generality, let $r_{2}>r_{1}$. Let $O_{1}, O_{2}$ be the centers of $\omega_{1}, \omega_{2}$ respectively. Consider a point $B_{0}$ such that $O_{1} O_{2} B_{0} A$ is an isosceles trapezoid with legs $O_{1} A=O_{2} B_{0}=r_{1}$ and diagonals $O_{1} B_{0}=A O_{2}=r_{2}$. Note that the circle with center $B_{0}$ and radius $r_{1}+r_{2}$ touches $\omega_{1}$ and $\omega_{2}$ externally. We will prove that $B=B_{0}$ and thus $r=r_{1}+r_{2}$. Suppose $B \neq B_{0}$. Then $B O_{2}+B_{0} O_{1} = B O_{2}+r_{2} = r = B O_{1}+r_{1} = B O_{1}+B_{0} O_{2}$, that is, in the quadrilateral (trapezoid) with vertices $O_{1}, O_{2}, B_{0}, B$, the sum of the diagonals equals the sum of the legs. This is impossible - a contradiction.