3 [Trigonometric ratios in a right-angled triangle]

Circles with centers $O 1$ and $O 2$ intersect at points A and $B$. It is known that $\angle A O 1 B=90^{\circ}, \angle A O 2 B=60^{\circ}$, $O 1 O 2=a$. Find the radii of the circles.
Let the line of centers $O_1O_2$ intersect the common chord $AB$ of the circles at point $M$. Then $M$ is the midpoint of $AB$ and $O_1O_2 \perp AB$. Triangle $AO_1B$ is a right isosceles triangle, and triangle $AO_2B$ is equilateral. Therefore, if $r$ and $R$ are the radii of the circles with centers $O_1$ and $O_2$ respectively, then $AB = r\sqrt{2}$ and $AB = R$, so $R = r\sqrt{2}$. Then,

$$
O_1M = \frac{r\sqrt{2}}{2}, \quad O_2M = \frac{R\sqrt{3}}{2} = \frac{r\sqrt{2} \cdot \sqrt{3}}{2} = \frac{r\sqrt{6}}{2}
$$

Assume that the centers of the circles lie on opposite sides of the line $AB$. Then $O_1M + MO_2 = O_1O_2$, or $\frac{r\sqrt{2}}{2} + \frac{r\sqrt{6}}{2} = a$. From this, we find that

$$
r = \frac{a\sqrt{2}}{\sqrt{3} + 1}, \quad R = r\sqrt{2} = \frac{2a}{\sqrt{3} + 1}.
$$

If the points $O_1$ and $O_2$ lie on the same side of the line $AB$, then $O_2M - MO_1 = O_1O_2$. Then

$$
r = \frac{a\sqrt{2}}{\sqrt{3} - 1}, \quad R = r\sqrt{2} = \frac{2a}{\sqrt{3} - 1}.
$$

## Answer

$$
\frac{a\sqrt{2}}{\sqrt{3} + 1}, \frac{2a}{\sqrt{3} + 1} \text{ or } \frac{a\sqrt{2}}{\sqrt{3} - 1}, \frac{2a}{\sqrt{3} - 1}.
$$