3. Inside the circle $\omega$ are located intersecting at points $K$ and $L$ circles $\omega_{1}$ and $\omega_{2}$, touching the circle $\omega$ at points $M$ and $N$. It turned out that points $K, M$, and $N$ lie on the same line. Find the radius of the circle $\omega$, if the radii of the circles $\omega_{1}$ and $\omega_{2}$ are 3 and 5, respectively.
Answer: 8.

![](https://cdn.mathpix.com/cropped/2024_05_06_66368caef28b3de8e403g-03.jpg?height=500&width=734&top_left_y=1569&top_left_x=727)

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

$$
\angle M K O_{1} = \angle K M O_{1} = \angle K N O_{2} = \angle N K O_{2}
$$

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

$$
M O = M O_{1} + O_{1} O = M O_{1} + K O_{2} = 3 + 5 = 8
$$