Isosceles triangle $ABC$ has $AB=AC=8$, and a circle of radius $12$ is tangent to line $AB$ at $B$ and line $AC$ at $C$. What is the area of the circle through $A,B,C$?

1. $D$ is the tangent circle's center, so $BD\perp AB$ and $CD\perp AC$. With $b=AC=AB,d=BD=$ tangent circle's radius: $e=AD=\sqrt{b^2+d^2}$ (right triangle $ABD$), and by the area formula (theorem 38) the circumradius of $\triangle ABC$ is $R=e/2$.
2. With $b=8,d=12$: $e=\sqrt{64+144}=\sqrt{208}=4\sqrt{13}$, so $R=2\sqrt{13}$.
3. Area $=\pi R^2=52\pi$, i.e. $\boxed{52\pi}$.