Drop a perpendicular from point $B$ to the chord $C E$. Apply the theorem about the angle between a tangent and a chord.

## Solution

Let $M$ be the projection of point $B$ onto the chord $C E$ of the second circle. Then $M$ is the midpoint of $C E$. Denote $\angle B C M=\alpha$. From the right triangle $BMC$, we find that

$$
\cos \alpha=\frac{C M}{B C}=\frac{\frac{3}{2}}{2}=\frac{3}{4}
$$

Then

$$
\sin \alpha=\sqrt{1-\frac{9}{16}}=\frac{\sqrt{7}}{4}
$$

From the theorem about the angle between a tangent and a chord, it follows that $\angle B A C=\angle B C M=\alpha$.

Let $R$ be the desired radius of the first circle. By the sine rule,

$$
R=\frac{B C}{2 \sin \angle B A C}=\frac{B C}{2 \sin \alpha}=\frac{2}{2 \cdot \frac{\sqrt{7}}{4}}=\frac{4}{\sqrt{7}}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_39480ed8b0d60d88e837g-45.jpg?height=472&width=608&top_left_y=650&top_left_x=725)

## Answer

$\frac{4}{\sqrt{7}}$.
Segment $A B$ is the diameter of a circle. A second circle with center at point $B$ has a radius of 2 and intersects the first circle at points $C$ and $D$. Chord $C E$ of the second circle is part of a tangent to the first circle and is equal to 3. Find the radius of the first circle.