Let the side lengths of squares $ABCD$ and $EFGH$ be $5$ and $9$, respectively, $I$ and $J$ be the intersections of the chord and the circle, $K$ be the intersection of $AH$ and $EF$, $L$ be the foot of $D$ on $GH$ and $a$ = $IE$ = $JF$. We have $AH^2$ = $AL^2 + HL^2 = AL^2 + ED^2 = 14^2 + 2^2 = 200$, or $AH = 10 \sqrt{2}$. Similarly, $AG^2 = AL^2 + GL^2 = 245$, or $AG = 7 \sqrt{5}$. It’s easily seen that the two triangles $ACK$ and $HEK$ are similar and we have $ \frac {AK}{HK} = \frac {KD}{KE} = \frac {AD}{HE} = \frac {5}{9}$, or $ \frac {AH - HK}{HK} $ = $ \frac {10\sqrt{2}}{HK} - 1 = \frac {5}{9}$, and $HK = \frac {45\sqrt{2}}{7}$, $AK = \frac {25\sqrt{2}}{7}$. Similarly, $ \frac{DE - KE}{KE} = \frac{2}{KE} - 1 = \frac{5}{9}$, and $KE = \frac {9}{7}$, $DK = \frac {5}{7}$. Now per the intersecting chord theorem $AK*HK = IK*JK = (a + KE)*(DK + DF + a)$, or $\frac {25\sqrt{2}}{7}*\frac {45\sqrt{2}}{7} = (a + \frac {9}{7})*(\frac {5}{7} + 7 + a)$, and we come up with the quadratic equation $a^2$ + $9a$ – $36$ = $0$ which has the solution $a = 3$ and thus $IJ = 15$. Now let’s find the radius of the circle. By applying the law of cosines to triangle $AGH$, we get $AG^2 = AH^2 + GH^2$ – $2AH*GH*cos \angle AHG$, or $245 = 200 + 81$ – $180\sqrt{2}cos \angle AHG$, or $cos \angle AHG = \frac{1}{5\sqrt{2}}$ and $sin \angle AHG = \frac{7}{5\sqrt{2}}$. From there, we have $ \frac{AG}{sin \angle AHG} = 2r$ where $r$ is the radius of the circle, or $r = \frac{5\sqrt{10}}{2}$. Finally, $ \frac{IJ}{r} = \frac{3}{5}\sqrt{10}$.