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From point K, located outside a circle with center at point $O$, two tangents $M K$ and $N K$ are drawn to this circle ($M$ and $N$ are the points of tangency). On the chord $M N$, a point $C$ is taken ($M C < C N$). Through point $C$, a line perpendicular to segment $OC$ is drawn, intersecting segment $N K$ at point $B$. It is known that the radius of the circle is $R$, $\angle M K N = \alpha$, and $M C = b$. Find $C B$.
Prove that $\angle C B O=\frac{\alpha}{2}$.

## Solution

Let $D$ be the point of intersection of segment $K O$ with chord $M N$. Then $D$ is the midpoint of $M N$, and since $\angle D N O=\angle O K N=\frac{\alpha}{2}$, we have

$$
\begin{gathered}
D N=O N \cos \angle D N O=R \cos \frac{\alpha}{2}, M N=2 D N=2 R \cos \frac{\alpha}{2} \\
C N=M N-M C=2 R \cos \frac{\alpha}{2}-b
\end{gathered}
$$

From triangle $C O N$, using the cosine rule, we find that

$$
\begin{gathered}
C O=\sqrt{N C^{2}+N O^{2}-2 N C \cdot N O \cos \frac{\alpha}{2}}= \\
=\sqrt{\left(2 R \cos \frac{\alpha}{2}-b\right)^{2}+R^{2}-2 R\left(2 R \cos \frac{\alpha}{2}-b\right) \cos \frac{\alpha}{2}}= \\
=\sqrt{R^{2}+b^{2}-2 R b \cos \frac{\alpha}{2}}
\end{gathered}
$$

Since points $O, C, B$, and $N$ lie on a circle with diameter $O B$, we have $\angle C B O=\angle C N O=\frac{\alpha}{2}$. Therefore,

$$
C B=C O \operatorname{ctg} \angle C B O=\operatorname{ctg} \frac{\alpha}{2} \sqrt{R^{2}+b^{2}-2 R b \cos \frac{\alpha}{2}}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_df2bbd76ac4dd2f0f20fg-23.jpg?height=469&width=978&top_left_y=1558&top_left_x=540)

## Answer

$\operatorname{ctg} \frac{\alpha}{2} \sqrt{R^{2}+b^{2}-2 R b \cos \frac{\alpha}{2}}$