jmc

Note that from the tangency condition that the supplement of $\angle CAB$ with respects to lines $AB$ and $AC$ are equal to $\angle AKB$ and $\angle AKC$, respectively, so from tangent-chord,$$\angle AKC=\angle AKB=180^{\circ }-\angle BAC$$Also note that $\angle ABK=\angle KAC$, so $△AKB\sim △CKA$. Using similarity ratios, we can easily find$$A{K}^2=BK\ast KC$$However, since $AB=7$ and $CA=9$, we can use similarity ratios to get$$BK=\frac{7}{9}AK,CK=\frac{9}{7}AK$$Now we use Law of Cosines on $△AKB$: From reverse Law of Cosines, $\cos \angle BAC=\frac{11}{21}⟹\cos (180^{\circ }-\angle BAC)=-\frac{11}{21}$. This gives us$$A{K}^2+\frac{49}{81}A{K}^2+\frac{22}{27}A{K}^2=49$$$$⟹\frac{196}{81}A{K}^2=49$$$$AK=\frac{9}{2}$$so our answer is $9+2=\boxed{11}$.