From point $M$ on the circle, three chords are drawn: $M N=1, M P=6, M Q=2$. The angles $N M P$ and $P M Q$ are equal. Find the radius of the circle.
Express the segments $N P$ and $P Q$ using the Law of Cosines from triangles $N M P$ and $P N Q$.

## Solution

Let $\angle N M P=\angle P M Q=\alpha$. We will express the equal segments $N P$ and $P Q$ using the Law of Cosines from triangles $N M P$ and $P M Q$ respectively:

$$
\begin{aligned}
& N P^{2}=M N^{2}+M P^{2}-2 N M \cdot M P \cos \alpha \\
& P Q^{2}=M P^{2}+M Q^{2}-2 M P \cdot M Q \cos \alpha
\end{aligned}
$$

By equating the right-hand sides of the obtained equations, we get an equation from which we find that $\cos \alpha=\frac{1}{4}$. Then

$$
\sin \alpha=\frac{\sqrt{15}}{4}
$$

If $R$ is the desired radius, then

$$
R=\frac{N P}{2 \sin \alpha}=\frac{\sqrt{34}}{\frac{\sqrt{15}}{2}}=2 \sqrt{\frac{34}{15}}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_ae1b0d3bbbf9aa5c38e8g-09.jpg?height=465&width=474&top_left_y=1231&top_left_x=792)

## Answer

$$
2 \sqrt{\frac{34}{15}}
$$