Answer: 282

Solution:

Both circles are tangent to the $x$-axis and to $y = mx,$ so both centers lie on the bisector of the first-quadrant angle between those lines. If the bisector makes angle $\alpha$ with the $x$-axis, then $m = \tan 2\alpha,$ and each center has the form $(x_i, \, x_i \tan\alpha)$ with radius $r_i = x_i \tan\alpha$ (its distance to the $x$-axis).

Since $(9, 6)$ lies on each circle, $(9 - x_i)^2 + (6 - x_i \tan\alpha)^2$ $= x_i^2 \tan^2\alpha,$ which expands to $$\begin{aligned} &x_i^2 - (18 + 12\tan\alpha)\,x_i + 117 \\ &= 0. \end{aligned}$$ Both $x_1$ and $x_2$ satisfy this one quadratic, so by Vieta's formulas $x_1 x_2 = 117.$ Then $r_1 r_2$ $= x_1 x_2 \tan^2\alpha$ $= 117 \tan^2\alpha$ $= 68,$ so $\tan^2\alpha = \frac{68}{117}$ and $\tan\alpha = \frac{2\sqrt{17}}{3\sqrt{13}}.$

Finally, $$\begin{aligned} m &= \frac{2\tan\alpha}{1 - \tan^2\alpha} \\ &= \frac{2\tan\alpha}{49/117} \\ &= \frac{234}{49} \cdot \frac{2\sqrt{17}}{3\sqrt{13}} \\ &= \frac{156\sqrt{17}}{49\sqrt{13}} \\ &= \frac{12\sqrt{221}}{49}, \end{aligned}$$ so $a + b + c$ $= 12 + 221 + 49$ $= 282.$