### 12.398 The distance between the centers of two externally tangent circles is $d$. The angle between their common external tangents is $\alpha$ radians. Find the area of the curvilinear triangle bounded by a segment of one tangent and the two corresponding arcs of the circles.
## Solution.

![](https://cdn.mathpix.com/cropped/2024_05_21_f024bf2ff7725246f3bfg-286.jpg?height=258&width=436&top_left_y=742&top_left_x=75)

Let $O$ and $P$ be the centers of two given circles with radii $x$ and $y$, touching at point $C$, and $AB$ be their external tangent. Let $PK$ be the height of the trapezoid $AOPB$. $\angle OPK = \frac{\alpha}{2}$ (by condition),
Since $x - y = OK = OP \cdot \sin \frac{\alpha}{2} = d \cdot \sin \frac{\alpha}{2}$, $PK = d \cdot \cos \frac{\alpha}{2}$, and $d = x + y$, we have

\[
\begin{gathered}
x^{2} - y^{2} = d^{2} \cdot \sin \frac{\alpha}{2} ; \quad x^{2} + y^{2} = \frac{(x + y)^{2} + (x - y)^{2}}{2} = \frac{d^{2}}{2} \left(1 + \sin^{2} \frac{\alpha}{2}\right) \\
S_{AOPB} = \frac{OA + PB}{2} \cdot PK = \frac{d^{2}}{2} \cdot \cos \frac{\alpha}{2}
\end{gathered}
\]

In the end, we have $\quad S_{ACB} = S_{AOPB} - (S_{OCA} + S_{PBC})$

Answer: $S_{ACB} = \frac{d^{2}}{8} \left(4 \cos \frac{\alpha}{2} - \pi \left(1 + \sin^{2} \frac{\alpha}{2}\right) + 2 \alpha \cdot \sin \frac{\alpha}{2}\right)$