Answer: 11

Solution:

By the tangent-chord angle in $\omega_1$ (tangent $AC,$ chord $AK$), $\angle KAC = \angle KBA,$ and in $\omega_2$ (tangent $AB,$ chord $AK$), $\angle KAB = \angle KCA.$ Write $u = \angle KAC$ and $v = \angle KAB,$ so $u + v = \angle A.$ Triangles $KAB$ and $KCA$ then have $\angle KAB = v = \angle KCA$ and $\angle KBA = u = \angle KAC,$ so $\triangle KAB \sim \triangle KCA.$ With $t = AK$ this gives $$\frac{KB}{t} = \frac{t}{KC} = \frac{AB}{CA} = \frac{7}{9},$$ so $KB = \frac{7t}{9}$ and $KC = \frac{9t}{7}.$ Also $\angle AKB = \angle AKC$ $= 180^\circ - u - v$ $= 180^\circ - \angle A,$ so $\angle BKC = 360^\circ - 2(180^\circ - \angle A)$ $= 2\angle A.$

From the law of cosines in $ABC,$ $\cos A = \frac{49 + 81 - 64}{2 \cdot 7 \cdot 9} = \frac{11}{21},$ so $\cos 2A = 2\left(\frac{11}{21}\right)^2 - 1 = -\frac{199}{441}.$ The law of cosines in triangle $BKC$ gives $$\begin{aligned} 64 &= \frac{49t^2}{81} + \frac{81t^2}{49} - 2t^2\cos 2A \\ &= t^2 \cdot \frac{2401 + 6561 + 3582}{3969} \\ &= \frac{12544\,t^2}{3969}, \end{aligned}$$ so $t^2 = \frac{64 \cdot 3969}{12544} = \frac{3969}{196}$ and $t = \frac{63}{14} = \frac{9}{2}.$

Hence $AK = \frac{9}{2}$ and $m + n = 9 + 2 = 11.$