2021 AMC Fall 10A Problem 15

Call the center of the given circle $O$. The given circle is not the one we want; it only touches $B$ and $C$, and does not pass through $A$.

1. Find right angles. A radius drawn to the point where a line touches a circle is perpendicular to that line. So $OB$ is perpendicular to $AB$, and $OC$ is perpendicular to $AC$. That makes $ABO$ and $ACO$ two right triangles, with right angles at $B$ and at $C$. They share the long side (hypotenuse) $AO$.
2. Find the center of the circle through $A$, $B$, $C$. Let $M$ be the midpoint of $AO$. In any right triangle, the midpoint of the long side is the same distance from all three corners. So:
- in triangle $ABO$: $MA = MB = MO$;
- in triangle $ACO$: $MA = MC = MO$.

So $M$ is the same distance from $A$, $B$ and $C$. It is the center of the circle through them, and the radius is half of $AO$.
3. Find $AO$. Use the Pythagorean theorem in triangle $ABO$:
$$AO^2 = AB^2 + OB^2 = (3\sqrt6)^2 + (5\sqrt2)^2 = 54 + 50 = 104.$$
4. Find the area. The radius is $R = \frac{AO}{2}$, so
$$R^2 = \frac{AO^2}{4} = \frac{104}{4} = 26.$$
The area is $\pi R^2 = 26\pi$.

The answer is $\boxed{\textbf{(C)}\ 26\pi}$.

Tangent lines give right angles for free. Two right triangles that share the same long side fit inside one circle, with that long side as a diameter. Seeing that circle turns the problem into one use of the Pythagorean theorem. Next time a picture has several right angles resting on the same segment, look for the circle that has that segment as its diameter.

Let $O$ be the center of the given circle. A radius drawn to a point of tangency is perpendicular to the tangent line, so $OB \perp AB$ and $OC \perp AC$: the angles $\angle ABO$ and $\angle ACO$ are both right angles.

A right angle subtended by segment $AO$ means that $B$ and $C$ both lie on the circle with diameter $AO$ (Thales' theorem in reverse). Since $A$ is trivially on that circle too, the circle with diameter $AO$ is the circle through $A$, $B$, $C$.

Compute $AO$ from right triangle $ABO$:
$$AO^2 = AB^2 + OB^2 = (3\sqrt6)^2 + (5\sqrt2)^2 = 54 + 50 = 104.$$
The circumradius is $R = \tfrac12 AO$, so $R^2 = \tfrac{104}{4} = 26$ and the area is $\pi R^2 = 26\pi$.

The answer is $\boxed{\textbf{(C)}\ 26\pi}$.