$\left.\begin{array}{l}\text { Tangent circles }\end{array}\right]$

Two circles touch each other internally at point $A$. From the center $O$ of the larger circle, a radius $O B$ is drawn, touching the smaller circle at point $C$. Find $\angle B A C$.
Express the desired angle in terms of angle $A O B$.

## Solution

Let $\angle A O B=\alpha$. Let $Q-$ be the center of the smaller circle. Since triangle $A O B$ is isosceles, then

$$
\angle O A B=\angle O B A=90^{\circ}-\frac{\alpha}{2}
$$

Since triangle $O Q C$ is a right triangle, then

$$
\angle O Q C=90^{\circ}-\alpha
$$

But $O Q C$ is the exterior angle of the isosceles triangle $A Q C$, so

$$
\angle Q A C=\frac{1}{2} \angle O Q C=45^{\circ}-\frac{\alpha}{2}
$$

Therefore,

$$
\angle B A C=\angle O A B-\angle Q A C=45^{\circ}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_5827e7e7a03e1b3b311bg-02.jpg?height=466&width=832&top_left_y=762&top_left_x=613)

## Answer

$45^{\circ}$.

## Problem