-Diagram by Brendanb4321

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 $\triangle AKB\sim \triangle CKA$. Using similarity ratios, we can easily find \[AK^2=BK*KC\] 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 $\triangle AKB$: From reverse Law of Cosines, $\cos{\angle BAC}=\frac{11}{21}\implies \cos{(180^{\circ}-\angle BAC)}=\angle AKB=-\frac{11}{21}$
Giving us \[AK^2+\frac{49}{81}AK^2+\frac{22}{27}AK^2=49\] \[\implies \frac{196}{81}AK^2=49\] \[AK=\frac{9}{2}\] so our answer is $9+2=\boxed{011}$.
$^{(*)}$ Let $O$ be the center of $\omega_1$. Then $\angle KAC = 90 - \angle OAK = 90 - \frac{1}{2}(180 - \angle AOK) = \frac{\angle AOK}{2} = \angle ABK$. Thus, $\angle ABK = \angle KAC$
-franchester; $^{(*)}$ by firebolt360

Supplement
In order to get to the Law of Cosines first, we first apply the LOC to $\triangle{ABC},$ obtaining $\angle{BAC}.$
We angle chase before applying the law of cosines to $\angle{AKB}.$
Note that $\angle{ABK}=\angle{KAC}$ and $\angle{KCA}=\angle{KAB}$ from tangent-chord. 
Thus, $\angle{AKC}=\angle{AKB}=180^{\circ}-(\angle{ABK}+\angle{KAB}).$
However from our tangent chord, note that:
\[\angle{ABK}+\angle{KAB}=\angle{KAC}+\angle{KAB}=\angle{BAC}.\]
Thus, $\angle{AKB}=180^\circ-\angle{BAC}.$

As an alternative approach, note that the sum of the angles in quadrilateral $ABKC$ is $360^{\circ}$ and we can find $\angle{AKB}=\frac12$ of convex $\angle{BKC},$ which is just:
\[\frac12 \left(360^{\circ}-2(\angle{KAB}+\angle{KBA}\right) = 180^\circ - \angle{BAC}.\]
~mathboy282
Triangle $ABC$ has side lengths $AB=7, BC=8,$ and $CA=9.$ Circle $\omega_1$ passes through $B$ and is tangent to line $AC$ at $A.$ Circle $\omega_2$ passes through $C$ and is tangent to line $AB$ at $A.$ Let $K$ be the intersection of circles $\omega_1$ and $\omega_2$ not equal to $A.$ Then $AK=\tfrac mn,$ where $m$ and $n$ are relatively prime positive integers. Find $m+n.$