smc

Let $O$ be the center of the circle, $\overline{AB}$ be the chord, and $M$ be the midpoint of $\overline{AB},$ as shown below. Note that $\overline{OM}\perp \overline{AB}.$ Since $OM=AM=BM=5,$ we conclude that $△OMA$ and $△OMB$ are congruent isosceles right triangles. It follows that $r=5\sqrt{2},$ so the area of $\odot O$ is $\pi r^2=\boxed{50\pi }$.