2021 Fall AMC 10A Problem 15

Let $O_1$ be the center of the circle tangent to $AB$ and $AC.$ Then $\angle ABO_1=\angle ACO_1=90^\circ,$ so $A,B,O_1,C$ are concyclic. Because the right angles at $B$ and $C$ subtend $AO_1,$ the segment $AO_1$ is a diameter of this circle. Let $O_2$ be its center. The same circle passes through $A,B,$ and $C,$ so it is the desired circumcircle. By the Pythagorean Theorem in $\triangle ABO_1,$ $$\begin{aligned} AO_1&=\sqrt{AB^2+BO_1^2}\\ &=\sqrt{54+50}=2\sqrt{26}. \end{aligned}$$ Therefore, the circumradius is $\sqrt{26},$ and the requested area is $26\pi.$ Thus, C is the correct answer.