3. Inside the circle $\omega$ is a circle $\omega_{1}$ that is tangent to it at point $K$. The circle $\omega_{2}$ is tangent to the circle $\omega_{1}$ at point $L$ and intersects the circle $\omega$ at points $M$ and $N$. It turns out that points $K, L$, and $M$ lie on the same line. Find the radius of the circle $\omega$, if the radii of the circles $\omega_{1}$ and $\omega_{2}$ are 4 and 7, respectively.
Answer: 11.

![](https://cdn.mathpix.com/cropped/2024_05_06_66368caef28b3de8e403g-07.jpg?height=474&width=879&top_left_y=1899&top_left_x=657)

Solution. Let $O, O_{1}, O_{2}$ be the centers of the circles $\omega, \omega_{1}, \omega_{2}$, respectively. The radii $O K$ and $O_{1} K$ of the circles $\omega$ and $\omega_{1}$ are perpendicular to their common tangent at point $K$. Therefore, they are parallel, which means that point $O_{1}$ lies on the segment $O K$ (see the figure). Since triangles $O K M$, $O_{1} K L$, and $O_{2} L M$ are isosceles,

$$
\angle K M O = \angle L K O_{1} = \angle K L O_{1} = \angle M L O_{2} = \angle L M O_{2}
$$

Therefore, $K O \parallel M O_{2}$ and $O_{1} O_{2} \parallel M O$, which means that $O O_{1} O_{2} M$ is a parallelogram. Then

$$
O M = O_{1} O_{2} = O_{1} L + L O_{2} = 4 + 7 = 11
$$