7. In the corners $C$ and $B$ of triangle $ABC$, circles with centers $O_{1}$ and $O_{2}$ of equal radius are inscribed, and point $O$ is the center of the circle inscribed in triangle $ABC$. These circles touch side $BC$ at points $K_{1}, K_{2}$, and $K$ respectively, with $C K_{1}=3, B K_{2}=7$, and $B C=16$.

a) Find the length of segment $C K$.

b) Let the circle with center $O_{1}$ touch side $AC$ at point $K_{3}$. Find the angle $ACB$, given that point $O_{1}$ is the center of the circle circumscribed around triangle $O K_{1} K_{3}$.
Answer. a) $C K=\frac{24}{5}$, b) $\angle A C B=2 \arcsin \frac{3}{5}=\arccos \frac{7}{25}$.

Solution. a) The lines $\mathrm{CO}_{1}$ and $\mathrm{BO}_{2}$ are the angle bisectors of angles $C$ and $B$ of the triangle, so they intersect at point $O$ - the center of the inscribed circle. Let the radii of the circles with centers $O_{1}$ and $O_{2}$ be $r$, and the radius of the inscribed circle be $R$. Triangles $O K B$ and $O_{2} K_{2} B$ are similar, with the similarity ratio equal to $\frac{R}{r}$, so $B K=\frac{7 R}{r}$. Similarly, $C K=\frac{3 R}{r}$, hence $\frac{10 R}{r}=16, C K=\frac{24}{5}$.

b) From the condition, it follows that $O_{1} O=O_{1} K_{1}=r$. Drop a perpendicular $O_{1} H$ from point $O_{1}$ to segment $O K$. Then $O H=R-r=\frac{3}{5} r, \angle O C B=\angle O O_{1} H=\arcsin \frac{O H}{O O_{1}}=\arcsin \frac{3}{5}$. Therefore, $\angle A C B=2 \arcsin \frac{3}{5}=\arccos \frac{7}{25}$.