3. Two circles $\omega_{1}$ and $\omega_{2}$ of the same radius $R$ intersect and are positioned such that the center of one circle is outside the other. Let the center of circle $\omega_{1}$ be denoted by $O_{1}$, and the center of circle $\omega_{2}$ by $O_{2}$. Suppose $B$ is one of the two intersection points of the circles. It turns out that the line $O_{1} B$ intersects the circle $\omega_{2}$ again at point $C$ such that $O_{2} C$ and $O_{1} O_{2}$ are perpendicular. Find the distance between the centers of the circles.

Answer: $R \sqrt{3}$.

Solution 1. Let $\angle O_{2} O_{1} C=\alpha$. Then $\angle O_{2} O_{1} B=\alpha$ and $\angle O_{1} O_{2} B=\alpha$ (since $O_{1} B=O_{2} B$). Therefore, $\angle \mathrm{BO}_{2} \mathrm{C}=90^{\circ}-\alpha=\angle \mathrm{BCO}_{2}$. Hence, $\mathrm{O}_{2} \mathrm{~B}=\mathrm{BC}=\mathrm{R}$, i.e., in the right triangle $\mathrm{O}_{1} \mathrm{O}_{2} \mathrm{C}$, the leg $\mathrm{O}_{2} \mathrm{C}=\mathrm{R}$, and the hypotenuse $\mathrm{O}_{1} \mathrm{C}=2 \mathrm{R}$. Therefore, the leg $\mathrm{O}_{1} \mathrm{O}_{2}=R \sqrt{3}$.
Solution 2. Let the second intersection point of the circles be denoted by the letter $A$. The circles are symmetric with respect to the line $A B$. Let the point of intersection of $A B$ and $\mathrm{O}_{1} \mathrm{O}_{2}$ be denoted by $D$. Due to symmetry, $A B$ and $\mathrm{O}_{1} \mathrm{O}_{2}$ are perpendicular, $\mathrm{O}_{1} \mathrm{D}=\mathrm{DO} \mathrm{O}_{2}$, and therefore, $\mathrm{DB}$ is the midline of triangle $\mathrm{O}_{1} \mathrm{O}_{2} \mathrm{C}$. Consequently, $D B=O_{2} C / 2=R / 2$. By the Pythagorean theorem, $\quad 0_{1} D^{2}=O_{1} B^{2}-D B^{2}=R^{2}-R^{2} / 4=3 R^{2} / 4$, so $O_{1} O_{2}=2 O_{1} D=2(R \sqrt{3} / 2)=R \sqrt{3}$.

Evaluation criteria.

It is established that $0_{1} C=2 R-3$ points.

It is established that D B (see solution 2) is the midline of triangle $\mathrm{O}_{1} \mathrm{O}_{2} \mathrm{C},-2$ points.