90. Two circles with radii $R$ and $r$ intersect, $A$ is one of the points of intersection. $B C$ is a common tangent ( $B$ and $C$ are the points of tangency). Find the radius of the circle circumscribed about triangle $A B C$.
90. Let (Fig. 21) $\widehat{O_{1} A O_{2}}=\varphi\left(O_{1}, O_{2}\right.$ be the centers of the circles, $A$ be the point of their intersection farthest from $B C$). We will show that $\widehat{B A C}=\frac{\varphi}{2} \cdot\left(\right.$ For the other point, the angle will be $\left.180^{\circ}-\frac{\varphi}{2} \cdot\right)$ Indeed,

$\widehat{B A C}=180^{\circ}-\widehat{A B C}-\widehat{B C A}=180^{\circ}-\left(90^{\circ}-\widehat{A B O_{1}}\right)-\left(90^{\circ}-\widehat{A C O_{2}}\right)=$

$$
=\widehat{A B O_{1}}+\widehat{A C O_{2}}=\widehat{B A O_{1}}+\widehat{C A O_{2}}=\widehat{O_{1} A O_{2}}-\widehat{B A C}=\varphi-\widehat{B A C}
$$

Let $\left|O_{1} O_{2}\right|=a$. By drawing $O_{2} M \| B C\left(M-\right.$ on $\left.O_{1} B\right)$, we get

$$
|B C|=\left|O_{2} M\right|=\sqrt{\left|O_{1} O_{2}\right|^{2}-\left|O_{1} M\right|}=\sqrt{a^{2}-(R-r)^{2}}
$$

From $\triangle O_{1} A O_{2}$, we find $\cos \varphi=\frac{R^{2}+r^{2}-a^{2}}{2 R r}$; thus, the radius of the circumcircle of $\triangle A B C$ is

$$
\frac{|B C|}{2 \sin \frac{\varphi}{2}}=\frac{\sqrt{a^{2}-(R-r)^{2}}}{2 \sqrt{\frac{1-\cos \varphi}{2}}}=\frac{\sqrt{a^{2}-(R-r)^{2}}}{\sqrt{2} \sqrt{1-\frac{R^{2}+r^{2}-a^{2}}{2 R r}}}=\sqrt{R_{r}}
$$

Answer: $\sqrt{\overline{R r}}$ (for both triangles).

![](https://cdn.mathpix.com/cropped/2024_05_21_6282cf4a54d2a9633c48g-086.jpg?height=307&width=439&top_left_y=1113&top_left_x=209)

Fig. 21.

![](https://cdn.mathpix.com/cropped/2024_05_21_6282cf4a54d2a9633c48g-086.jpg?height=301&width=323&top_left_y=1118&top_left_x=701)

Fig. 22.