jmc

Let the triangle be $ABC$, with hypotenuse $\overline{AB}$, and let $O$ be the center of the circle. The hypotenuse of a right triangle that is inscribed in a circle is a diameter of the circle, so $\overline{AB}$ is the diameter of the circle. Since point $C$ is on the circle, point $C$ is $100/2=50$ units from the midpoint of $\overline{AB}$ (which is the center of the circle). So, point $C$ cannot be any more than 50 units from $\overline{AB}$. This maximum can be achieved when $\overline{OC}\perp \overline{AB}$. The area of $△ABC$ then is $(50)(100)/2=\boxed{2500}$ square units.