7. Circles with centers \(O_{1}\) and \(O_{2}\) of equal radius are inscribed in the corners \(A\) and \(B\) of triangle \(ABC\), respectively, and point \(O\) is the center of the circle inscribed in triangle \(ABC\). These circles touch side \(AB\) at points \(K_{1}\), \(K_{2}\), and \(K\) respectively, with \(AK_{1}=4\), \(BK_{2}=6\), and \(AB=16\).

a) Find the length of segment \(AK\).

b) Let the circle with center \(O_{1}\) touch side \(AC\) at point \(K_{3}\). Find the angle \(CAB\) if it is known that point \(O_{1}\) is the center of the circle circumscribed around triangle \(OK_{1}K_{3}\).
Answer. a) $A K=\frac{32}{5}$, b) $\angle C A B=2 \arcsin \frac{3}{5}=\arccos \frac{7}{25}$.

Solution. a) The lines $A O_{1}$ and $B O_{2}$ are the angle bisectors of angles $A$ 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 $B K=\frac{6 R}{r}$. Similarly, $A K=\frac{4 R}{r}$, hence $\frac{10 R}{r}=16, A K=\frac{32}{5}$.

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 $O H=R-r=\frac{3}{5} r, \angle O A B=\angle O O_{1} H=\arcsin \frac{O H}{O O_{1}}=\arcsin \frac{3}{5}$. Therefore, $\angle C A B=2 \arcsin \frac{3}{5}=\arccos \frac{7}{25}$.