smc

Let $O$ be the center of the circle, and $X$ be the midpoint of $\overline{CD}$. Let $CX=a$ and $EX=b$. This implies that $DE=a-b$. Since $CE=CX+EX=a+b$, we now want to find $(a+b{)}^2+(a-b{)}^2=2(a^2+b^2)$. Since $\angle CXO$ is a right angle, by Pythagorean theorem $a^2+b^2=C{X}^2+O{X}^2=(5\sqrt{2}{)}^2=50$. Thus, our answer is $2\times 50=\boxed{100}$.