jmc

Let $B=(0,0)$, $C=(s,0)$, $A=(0,s)$, $D=(s,s)$, and $E=\left(s+\frac{r}{\sqrt{2}},s+\frac{r}{\sqrt{2}}\right)$. Apply the Pythagorean Theorem to $△AFE$ to obtain $$r^2+\left(9+5\sqrt{2}\right)={\left(s+\frac{r}{\sqrt{2}}\right)}^2+{\left(\frac{r}{\sqrt{2}}\right)}^2,$$from which $9+5\sqrt{2}=s^2+rs\sqrt{2}$. Because $r$ and $s$ are rational, it follows that $s^2=9$ and $rs=5$, so $r/s=\boxed{\frac{5}{9}}$.

OR

Extend $\overline{AD}$ past $D$ to meet the circle at $G\ne D$. Because $E$ is collinear with $B$ and $D$, $△EDG$ is an isosceles right triangle. Thus $DG=r\sqrt{2}$. By the Power of a Point Theorem, $$\begin{array}{rl}9+5\sqrt{2}& =A{F}^2\\ & =AD\cdot AG\\ & =AD\cdot \left(AD+DG\right)\\ & =s\left(s+r\sqrt{2}\right)\\ & =s^2+rs\sqrt{2}.\end{array}$$As in the first solution, we conclude that $r/s=\boxed{\frac{5}{9}}$.