10.286. Two circles of radii $r$ and $3r$ touch each other externally. Find the area of the figure enclosed between the circles and their common external tangent.
## Solution.

Let $O_{1}$ and $O_{2}$ be the centers of circles with radii $r$ and $3r$ respectively (Fig. 10.79), $AB$ be their common external tangent, $A$ and $B$ be the points of tangency. Then $O_{1}A \perp AB, O_{2}B \perp AB, O_{1}A = r, O_{2}B = 3r, O_{1}O_{2} = 4r$. $P$ is the point of tangency of the given circles. Then the desired area

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0769.jpg?height=452&width=570&top_left_y=77&top_left_x=143)

Fig. 10.79

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0769.jpg?height=476&width=486&top_left_y=67&top_left_x=724)

Fig. 10.80

$S = S_{1} - (S_{2} + S_{3})$, where $S_{1}$ is the area of trapezoid $O_{1}ABO_{2}$, $S_{2}$ is the area of sector $AO_{1}P$, and $S_{3}$ is the area of sector $BO_{2}P$. $O_{1}C$ is the height of trapezoid $O_{1}ABO_{2}$. Therefore, $BC = AO_{1} = r$, $CO_{2} = BO_{2} - BC = 2r$. Hence, $\angle CO_{1}O_{2} = 30^{\circ}$, $\angle CO_{2}O_{1} = 60^{\circ}$, $O_{1}C = 2r\sqrt{3}$, $\angle AO_{1}O_{2} = 120^{\circ}$. $S_{1} = \frac{AO_{1} + BO_{2}}{2} \cdot O_{1}C = 4r^{2}\sqrt{3}$, $S_{2} = \frac{\pi r^{2} \cdot 120}{360} = \frac{\pi r^{2}}{3}$, $S_{3} = \frac{\pi (3r)^{2} \cdot 60}{360} = \frac{3\pi r^{2}}{2}$. Then $S = 4r^{2}\sqrt{3} - \left(\frac{\pi r^{2}}{3} + \frac{3\pi r^{2}}{2}\right) = \frac{r^{2}(24\sqrt{3} - 11\pi)}{6}$.

Answer: $\frac{r^{2}(24\sqrt{3} - 11\pi)}{6}$.