Square $ABCD$ has side length $s$, a circle centered at $E$ has radius $r$, and $r$ and $s$ are both rational. The circle passes through $D$, and $D$ lies on $\overline{BE}$. Point $F$ lies on the circle, on the same side of $\overline{BE}$ as $A$. Segment $AF$ is tangent to the circle, and $AF=\sqrt{9+5\sqrt{2}}$. What is $r/s$?

[asy]
pair A,B,C,D,I,F;
A=(0,10); B=(0,0); C=(10,0); D=(10,10);

I=(14,13); F=(11,17);
draw(A--B--C--D--cycle,linewidth(0.7));
draw(Circle(I,5),linewidth(0.7));
draw(A--F,linewidth(0.7));
label("$A$",A,NW);
label("$B$",B,SW);
label("$C$",C,SE);
label("$D$",D,SW);
label("$F$",F,N);
label("$E$",I,E);
dot(I);
[/asy]
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 $\triangle AFE$ 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 $D$, $\triangle EDG$ is an isosceles right triangle.  Thus $DG = r\sqrt{2}$.  By the Power of a Point Theorem, \begin{align*}
9+5\sqrt{2} &= AF^2 \\
&= AD\cdot AG\\
& = AD\cdot \left(AD+DG\right) \\
&=
s\left(s+r\sqrt{2}\right) \\
&= s^2+rs\sqrt{2}.\end{align*}As in the first solution, we conclude that $r/s=\boxed{\frac{5}{9}}$.