By power of a point, $(C E)(E D)=(A E)(E B)=16$, and $C E+E D=C D=$ 10. Thus $C E, E D$ are 2, 8. Without loss of generality, assume $C E=8$ and $D E=2$. Assume our circle is centered at the origin, with points $A=(-3,5), B=(-3,-5), C=(5,-3)$, $D=(-5,-3)$, and the equation of the circle is $x^{2}+y^{2}=34$. Clearly the largest possible rectangle must lie in the first quadrant, and if we let $(x, y)$ be the upper-right corner of the rectangle, then the area of the rectangle is $(x+3)(y+3)=9+6(x+y)+x y \leq 9+12 \sqrt{\frac{x^{2}+y^{2}}{2}}+\frac{x^{2}+y^{2}}{2}=26+6 \sqrt{17}$, where equality holds if and only if $x=y=\sqrt{17}$.
Chords $\overline{A B}$ and $\overline{C D}$ of circle $\omega$ intersect at $E$ such that $A E=8, B E=2, C D=10$, and $\angle A E C=90^{\circ}$. Let $R$ be a rectangle inside $\omega$ with sides parallel to $\overline{A B}$ and $\overline{C D}$, such that no point in the interior of $R$ lies on $\overline{A B}, \overline{C D}$, or the boundary of $\omega$. What is the maximum possible area of $R$?