imc

Let $\odot O_1$ be the circle with radius $5\sqrt{2}$ that is tangent to $\overleftrightarrow{AB}$ at $B$ and to $\overleftrightarrow{AC}$ at $C.$ Note that $\angle AB{O}_1=\angle AC{O}_1=90^{\circ }.$ Since the opposite angles of quadrilateral $AB{O}_{1}C$ are supplementary, quadrilateral $AB{O}_{1}C$ is cyclic. Let $\odot O_2$ be the circumcircle of quadrilateral $AB{O}_{1}C.$ It follows that $\odot O_2$ is also the circumcircle of $△ABC,$ as shown below: By the Inscribed Angle Theorem, we conclude that $\overline{A{O}_1}$ is the diameter of $\odot O_2.$ By the Pythagorean Theorem on right $△AB{O}_1,$ we have $$A{O}_1=\sqrt{A{B}^2+B{O}_1^2}=2\sqrt{26}.$$ Therefore, the area of $\odot O_2$ is $\pi \cdot {\left(\frac{A{O}_1}{2}\right)}^2=\boxed{26\pi }.$