Two circles of different radii touch at point $C$ of a straight line and are located on the same side of it. Segment $C D$ is the diameter of the larger circle. Two lines are drawn from point $D$, touching the smaller circle at points $A$ and $B$. The line passing through points $C$ and $A$ forms a $75^{\circ}$ angle with the common tangent to the circles at point $C$ and intersects the larger circle at point $M$. It is known that $A M=\sqrt{2-\sqrt{3}}$.

Find the area of the figure bounded by the tangent segments $D A, D B$ and the arc $A C B$ of the smaller circle.
Let $r$ and $R$ be the radii of the circles ($r < R$), $O_{1}$ and $O_{2}$ their centers, $E$ the second intersection point of line $CD$ with the smaller circle, and $P$ the projection of point $E$ onto $MD$. Note that $\angle AEC = \angle MDC = 75^{\circ}$.

Since $\angle CAE = \angle CMD = 90^{\circ}$, $AMPE$ is a rectangle. Therefore, $PE = MA = \sqrt{2 - \sqrt{3}} = \frac{\sqrt{3} - 1}{\sqrt{2}}$. From the right triangle $EPD$, we find that $PE = ED \sin 75^{\circ}$, or $\frac{\sqrt{3} - 1}{\sqrt{2}} = \frac{2(R - r)(\sqrt{3} + 1) \sqrt{2}}{4}$ (see problem $\underline{61207}$ a). Therefore, $R - r = 2 - \sqrt{3}$.

In the right triangle $O_{2}AD$, $AO_{2} = r$, $O_{2}D = 2R - r$, and $\angle AO_{2}D = 2 \angle ACO_{2} = 30^{\circ}$. Therefore, $r = \frac{(2R - r) \sqrt{3}}{2} = \frac{(4 - 2\sqrt{3} + r) \sqrt{3}}{2}$, from which we get $r = 2\sqrt{3}$.

The desired area is the sum of the areas of sector $O_{2}ACB \left(\frac{5 \pi r^{2}}{6}\right)$ and quadrilateral $O_{2}ADB \left(O_{2}A \cdot AD\right)$, which is $\frac{5 \pi r^{2}}{6} + \frac{r^{2}}{\sqrt{3}} = 2(5 \pi + 4 \sqrt{3})$.

![](https://cdn.mathpix.com/cropped/2024_05_06_3ae27f1f0c68451d31a7g-14.jpg?height=612&width=677&top_left_y=0&top_left_x=696)

## Answer

$2(5 \pi + 4 \sqrt{3})$

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