Point $O$ divides the segment $A B$ into segments $O A=6$ and $O B=4$. A circle is drawn with center at point $O$, and tangents are drawn from $A$ and $B$ to the circle, intersecting at point $M$, with the points of tangency lying on the same side of the line $A B$. Find the radius of the circle if $O M=12$.
Apply the Law of Cosines to triangles AMO and BMO.

## Solution

Since $MO$ is the bisector of angle $AMB$, then

$$
\frac{AM}{MB}=\frac{AO}{OB}=\frac{3}{2}
$$

Let $AM=3x, BM=2x, \angle AMB=2\varphi$. By the Law of Cosines from triangles $AMO$ and $BMO$, we find that

$$
\begin{aligned}
& AO^2=AM^2+MO^2-2AM \cdot MO \cos \varphi \\
& BO^2=MB^2+MO^2-2MB \cdot MO \cos \varphi
\end{aligned}
$$

or

$$
36=9x^2+144-72x \cos \varphi, 16=4x^2+144-48x \cos \varphi
$$

or

$$
24x \cos \varphi=3x^2+36, 24x \cos \varphi=2x^2+64
$$

Subtracting the obtained equations term by term, we get that $x^2=28$. Then

$$
\cos \varphi=\frac{5}{2\sqrt{7}}, \sin \varphi=\frac{\sqrt{3}}{2\sqrt{7}}
$$

Let $D$ be the point of tangency of the given circle with side $BM$. Then

$$
OD=OM \sin \varphi=\frac{12\sqrt{3}}{2\sqrt{7}}=\frac{6\sqrt{21}}{7}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_6ed8f1205731b80a0039g-28.jpg?height=395&width=829&top_left_y=174&top_left_x=617)

## Answer

$\frac{6\sqrt{21}}{7}$.