Answer: 17

Solution:

Let the radius-$5$ circles have centers $P_1$ and $P_2,$ so $P_1P_2 = 5 + 5 = 10,$ and let $M$ be the midpoint. The radius-$8$ circle's center $Q$ satisfies $QP_1 = QP_2 = 13,$ so $Q$ lies on the perpendicular bisector of $\overline{P_1P_2}$ at distance $\sqrt{13^2 - 5^2} = 12$ from $M.$ Likewise the fourth circle, of radius $r,$ has its center $R$ on the same perpendicular bisector with $RP_1 = 5 + r,$ so $RM = \sqrt{(5+r)^2 - 25}$ $= \sqrt{r^2 + 10r}.$

The small circle nestles in the space between the other three, so $R$ is between $M$ and $Q,$ and external tangency to the radius-$8$ circle gives $$12 - \sqrt{r^2 + 10r} = 8 + r.$$ Then $\sqrt{r^2 + 10r} = 4 - r,$ and squaring yields $r^2 + 10r = 16 - 8r + r^2,$ so $18r = 16$ and $r = \frac{8}{9}.$

Thus $m + n = 8 + 9 = 17.$