10.290. Two circles of radii $R$ and $r$ touch each other externally. A common external tangent is drawn to these circles, and a circle is inscribed in the curvilinear triangle thus formed. Find its area.
## Solution.

$A$ and $B$ are the points of tangency of a common external tangent with the touching circles with centers $O_{1}$ and radius $R$, and $O_{2}$ and radius $r$, respectively, where $R > r$ (Fig. 10.83). Then $O_{1} O_{2} = R + r$, $O_{1} A \perp A B$, $O_{1} A = R$, $O_{2} B = r$, $O_{1} A \parallel O_{2} B$. $O_{3}$ is the center of the circle whose area is sought in the problem. Through points $O_{2}$ and $O_{3}$, draw lines $M N$ and $O_{2} K$ parallel to $A B$. Then $O_{2} K \perp O_{1} A$, $M N \perp O_{1} A_{1}$, $O_{1} K = R - r$. $D$ is the point of tangency of the circle with center $O_{3}$ and $A B$. Then $O_{3} D \perp A B$, $O_{3} D = x$. $O_{1} M = O_{1} A - A M = R - x$, $O_{2} O_{3} = r + x$, $O_{1} O_{3} = R + x$, $O_{2} N = r - x$. In $\Delta O_{1} M O_{3}$ ($\angle O_{1} M O_{3} = 90^{\circ}$): $O_{3} M = \sqrt{O_{1} O_{3}^{2} - O_{1} M^{2}} = \sqrt{(R + x)^{2} - (R - x)^{2}} = 2 \sqrt{R x}$. In $\Delta O_{2} N O_{3}$ ($\angle O_{2} N O_{3} = 90^{\circ}$): $O_{3} N = \sqrt{O_{2} O_{3}^{2} - O_{2} N^{2}} = \sqrt{(r + x)^{2} - (r - x)^{2}} = 2 \sqrt{r x}$.

In $\Delta O_{1} K O_{2}$ ($\angle O_{1} K O_{2} = 90^{\circ}$): $O_{2} K = \sqrt{O_{1} O_{2}^{2} - O K^{2}} = \sqrt{(R + r)^{2} - (R - r)^{2}} = 2 \sqrt{R r}$. Since $K O_{2} = M N = M O_{3} + O_{3} N$, then $2 \sqrt{R r} = 2 \sqrt{R x} + 2 \sqrt{r x}$;

$\sqrt{x} = \frac{\sqrt{R r}}{\sqrt{R} + \sqrt{r}}$; $x = \frac{R r}{(\sqrt{R} + \sqrt{r})^{2}}$.

The area of the circle sought is $S = \pi x^{2} = \frac{\pi R^{2} r^{2}}{(\sqrt{R} + \sqrt{r})^{4}}$. Answer: $\frac{\pi R^{2} r^{2}}{(\sqrt{R} + \sqrt{r})^{4}}$.

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0772.jpg?height=446&width=680&top_left_y=629&top_left_x=338)

Fig. 10.84