6. Point $K$ lies on side $A B$ of triangle $A B C$ with an angle of $120^{\circ}$ at vertex $C$. Circles are inscribed in triangles $A K C$ and $B K C$ with centers $O$ and $Q$ respectively. Find the radius of the circumcircle of triangle $O Q C$, if $O K=6, K Q=7$.
Answer: $\sqrt{\frac{85}{3}}$.

Solution. The center of the circle inscribed in an angle lies on the bisector of this angle, so rays $K O$ and $K Q$ are the bisectors of angles $A K C$ and $B K C$. Since the angle between the bisectors of adjacent angles is a right angle, $\angle O K Q=90^{\circ}$, and then by the Pythagorean theorem, we find that $O Q=\sqrt{O K^{2}+K Q^{2}}=\sqrt{85}$.

Since $C O$ and $C Q$ are the bisectors of angles $A C K$ and $B C K$, then

$$
\angle O C Q=\angle O C K+\angle Q C K=\frac{1}{2} \angle A C K+\frac{1}{2} \angle B C K=\frac{1}{2} \angle B C A=60^{\circ} .
$$

By the generalized sine theorem for triangle $A D T$, we find that the desired radius is $\frac{O Q}{2 \sin \angle O C Q}=\sqrt{\frac{85}{3}}$.