## Task A-2.4.

Two circles that do not intersect are given, with radii $r_{1}$ and $r_{2}$. The distance between the points of tangency of the common internal tangent to these circles is 12, and the distance between the points of tangency of the common external tangent to these circles is 16. Determine the product $r_{1} r_{2}$.

The internal tangent (is the common tangent that) intersects the line segment joining the centers of the circles.
## First Solution.

Let points $A$ and $B$ be the points of tangency of the external tangent, and points $P$ and $Q$ be the points of tangency of the internal tangent with the given circles of radii $r_{1}$ and $r_{2}$, respectively. Let $S_{1}$ be the center of the circle with radius $r_{1}$ and $S_{2}$ be the center of the circle with radius $r_{2}$, and let $X$ be the intersection of the lines $A B$ and $P Q$.

![](https://cdn.mathpix.com/cropped/2024_05_30_9a2ee33c697c48e6349cg-13.jpg?height=525&width=1008&top_left_y=2059&top_left_x=453)

Consider triangles $A X S_{1}$ and $P X S_{1}$. Since $\varangle S_{1} A X=90^{\circ}=\varangle S_{1} P X$, $|S_{1} A| = r_{1} = |S_{1} P|$, and the side $\overline{X S_{1}}$ is common, by the S-S-K congruence theorem (side $\overline{X S_{1}}$ is the largest side, acting as the hypotenuse in the observed triangles), we conclude that triangles $A X S_{1}$ and $P X S_{1}$ are congruent.

From the obtained congruence, it follows that $|A X| = |P X|$ and $\varangle A X S_{1} = \varangle P X S_{1} = \frac{1}{2} \varangle A X P$.

Similarly, we obtain that triangles $B X S_{2}$ and $Q X S_{2}$ are congruent, from which it follows that $|B X| = |Q X|$ and $\varangle B S_{2} X = \varangle Q S_{2} X = \frac{1}{2} \varangle Q S_{2} B$.

Using the obtained equalities, we conclude

$$
16 = |A B| = |A X| + |X B| = |P X| + |Q X| = |P Q| + 2|Q X| = 12 + 2|Q X| \text{.}
$$

From the obtained equality, it follows that $|Q X| = 2$, from which we get $|X B| = 2$ and $|A X| = 14$.

Consider triangles $A X S_{1}$ and $B S_{2} X$. Since $\varangle S_{1} A X = 90^{\circ} = \varangle X B S_{2}$ and

$$
\begin{aligned}
& \varangle A X S_{1} = \frac{1}{2} \varangle A X P \stackrel{\text{ angles }}{=} \frac{1}{2}\left(180^{\circ} - \varangle B X Q\right) \text{ sum of angles in quadrilateral } B X Q S_{2} \\
& \frac{1}{2}\left(180^{\circ} - \left(360^{\circ} - \varangle X Q S_{2} - \varangle Q S_{2} B - \varangle S_{2} B X\right)\right) \stackrel{\text{ tangent is perpendicular to radius }}{=} \\
& \frac{1}{2}\left(180^{\circ} - \left(360^{\circ} - 90^{\circ} - \varangle Q S_{2} B - 90^{\circ}\right)\right) = \frac{1}{2} \varangle Q S_{2} B = \varangle B S_{2} X,
\end{aligned}
$$

by the K-K similarity theorem, we conclude that triangles $A X S_{1}$ and $B S_{2} X$ are similar.

From the obtained similarity, it follows that

$$
\frac{r_{1}}{|A X|} = \frac{|X B|}{r_{2}}
$$

Using what was previously shown, we get $r_{1} \cdot r_{2} = 2 \cdot 14 = 28$.