2. Circles $k(S ; 6 \mathrm{~cm})$ and $l(O ; 4 \mathrm{~cm})$ have an internal tangency at point $B$. Determine the lengths of the sides of triangle $A B C$, where point $A$ is the intersection of line $O B$ with circle $k$ and point $C$ is the intersection of circle $k$ with the tangent from point $A$ to circle $l$.
2. Let the point of tangency of the circle $l$ with the tangent from point $A$ be $D$ (Fig. 1). From the properties of the tangent to a circle, it follows that angle $A D O$ is a right angle. Similarly, angle $A C B$ is also a right angle (Thales' theorem).

![](https://cdn.mathpix.com/cropped/2024_04_17_e9e73f8b95c2cf108172g-3.jpg?height=611&width=700&top_left_y=394&top_left_x=678)

Fig. 1

Triangles $A B C$ and $A O D$ are similar by the AA similarity criterion, as they share angles $A C B$, $A D O$, and the common angle at vertex $A$. From this similarity, it follows that

$$
\frac{|B C|}{|O D|}=\frac{|A B|}{|A O|}
$$

From the given numerical values, we have $|O D|=|O B|=4 \mathrm{~cm}$, $|O S|=|S B|-|O B|=2 \mathrm{~cm}$, $|O A|=|O S|+|S A|=8 \mathrm{~cm}$, and $|A B|=12 \mathrm{~cm}$. According to (1), we have $|B C|: 4 \mathrm{~cm}=12: 8$, and thus $|B C|=6 \mathrm{~cm}$. Using the Pythagorean theorem for triangle $A B C$, we finally determine that $|A C|=\sqrt{12^{2}-6^{2}} \mathrm{~cm}=6 \sqrt{3} \mathrm{~cm}$.

For a complete and correctly justified solution, award 6 points. Of these, 2 points for the diagram and justification of the right angles, 2 points for the justification of the similarity of the triangles and the formulation of the necessary equation, and 2 points for the calculation of the lengths of sides $B C$ and $A C$.