10.045. The common chord of two intersecting circles is seen from the centers at angles of $90^{\circ}$ and $60^{\circ}$. Find the radii of the circles if the distance between their centers is $\sqrt{3}+1$.
Solution.

Let $r$ be the radius of one circle, and $R$ be the radius of the other circle. According to the problem, $\angle A O_{1} B=90^{\circ}, \angle A O_{2} B=60^{\circ}, O_{1} O_{2}=\sqrt{3}+1 ; A B=r \sqrt{2} ;$ $O_{1} E=A B / 2 ; O_{1} E=r \frac{\sqrt{2}}{2}, A B=2 R \cdot \sin 30^{\circ}=R\left(\angle A O_{2} E=\frac{\angle A O_{2} B}{2}\right)$, therefore, $R=r \sqrt{2}, E O_{2}=R \frac{\sqrt{3}}{2}=r \frac{\sqrt{6}}{2}, O_{1} O_{2}=O_{1} E+O_{2} E=\frac{r \sqrt{6}}{2}+\frac{r \sqrt{2}}{2}=$ $=\sqrt{3}+1, r(\sqrt{6}+\sqrt{2})=(\sqrt{3}+1) \cdot 2, r=\sqrt{2} ; R=\sqrt{2} \cdot \sqrt{2}=2$ (Fig. 10.45).

Answer: 2 and $\sqrt{2}$.