Square $ABCD$ has side length $s$, a circle centered at $E$ has radius $r$, and $r,s$ are 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 $A$, with $\overline{AF}$ tangent to the circle and $AF=\sqrt{16+6\sqrt2}$. What is $r/s$?

1. $AF^2=s^2+rs\sqrt2$ (from $AE^2=AF^2+r^2$ and the law of cosines in $\triangle ADE$ with $\angle ADE=135^\circ$).
2. Matching $AF^2=16+6\sqrt2$: $s^2=16,\ rs\sqrt2=6\sqrt2\Rightarrow rs=6$, so $s=4,r=\frac32$.
3. $r/s=\boxed{\dfrac38}$.