$\left[\begin{array}{l}\text { Common tangent to two circles }\end{array}\right]$
$\left[\begin{array}{l}\text { Area of a quadrilateral }\end{array}\right]$
$\left[\begin{array}{l}\text { Auxiliary similar triangles }\end{array}\right]$
[ Application of trigonometric formulas (geometry).]

Two circles with radii $R$ and $r$ are located outside each other. The segments of the common internal tangents $A C$ and $B D$ ($A, B, C, D$ - points of tangency) are equal to $a$. Find the area of the quadrilateral $A B C D$.
Find the sine of the angle between the diagonals of quadrilateral $ABCD$.

## Solution

Let $O_{1}$ and $O_{2}$ be the centers of circles with radii $r$ and $R$ respectively, points $A$ and $B$ belong to the first circle, $C$ and $D$ belong to the second, and $P$ is the intersection point of $AC$ and $BD$.

From the similarity of triangles $DPO_{2}$ and $BPO_{1}$, it follows that $DP: PB = R: r$. Therefore, $PD = R a / (R + r)$. Consequently, $\operatorname{tg} \angle DPO_{2} = D_{2} / PD = (R + r) / a$, and $\sin \angle DPC = \sin 2 \angle DPO_{2} = \frac{2 \frac{R+r}{a}}{1 + \left(\frac{R+r}{a}\right)^{2}} = \frac{2 a (R + r)}{a^{2} + (R + r)^{2}}$.

Therefore, $S_{ABCD} = \frac{1}{2} AC \cdot BD \sin \angle DPC = \frac{a^{3} (R + r)}{a^{2} + (R + r)^{2}}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_3ae27f1f0c68451d31a7g-05.jpg?height=460&width=761&top_left_y=1862&top_left_x=657)

Answer

$\frac{a^{3} (R + r)}{a^{2} + (R + r)^{2}}$