Answer: 104

Small Hint:

Put line $DF$ on the $x$-axis with $D$ at the origin, and write the coordinates of $A,$ $D,$ $H,$ and $G$ in terms of $DE$

Big Hint:

The circle’s center lies on $y = -8$ (bisector of $\overline{AD}$) and on the vertical bisector of $\overline{HG};$ equate its distances to $D$ and $H$

Solution:

Put line $DF$ on the $x$-axis with $D = (0, 0)$ and $C = (107, 0),$ so $A = (0, -16).$ Let $DE = e.$ Then $E = (e, 0),$ $F = (e + 184, 0),$ and the second rectangle sits above the line: $H = (e, 17)$ and $G = (e + 184, 17).$

The center of the circle through $A,$ $D,$ $H,$ $G$ lies on the perpendicular bisector of the vertical segment $\overline{AD},$ the line $y = -8,$ and on the perpendicular bisector of the horizontal segment $\overline{HG},$ the line $x = e + 92.$ Equating the center’s squared distances to $D$ and to $H,$ $$\begin{aligned} &(e + 92)^2 + 8^2 \\ &= 92^2 + 25^2 = 9089, \end{aligned}$$ so $(e + 92)^2 = 9025$ and $e + 92 = 95,$ giving $e = 3.$

Therefore $CE = DC - DE$ $= 107 - 3 = 104.$