2019 AMC 10B Problem 23

Let the tangents meet at $P = (p, 0)$. Tangent segments from an external point are equal, so $PA = PB$:
$$(p - 6)^2 + 13^2 = (p - 12)^2 + 11^2 \;\Longrightarrow\; -12p + 205 = -24p + 265 \;\Longrightarrow\; p = 5.$$
So $P = (5, 0)$ and $PA^2 = 1 + 169 = 170$.

Let $M = (9, 12)$ be the midpoint of $AB$ and $O$ the center. Both $O$ and $P$ are equidistant from $A$ and $B$, so $O$, $M$, $P$ are collinear on the perpendicular bisector of $AB$, and $AB \perp OP$.

Now use the right triangle $OAP$ (right angle at $A$, since a radius is perpendicular to a tangent). $AM$ is its altitude to the hypotenuse $OP$, so $AM^2 = PM \cdot MO$. Compute
$$AM = \tfrac12 AB = \tfrac12\sqrt{36 + 4} = \sqrt{10}, \qquad PM = \sqrt{(9-5)^2 + 12^2} = \sqrt{160} = 4\sqrt{10}.$$
Then $MO = \frac{AM^2}{PM} = \frac{10}{4\sqrt{10}} = \frac{\sqrt{10}}{4}$, and
$$r^2 = OA^2 = AM^2 + MO^2 = 10 + \frac{10}{16} = \frac{85}{8}.$$

The area is $\pi r^2 = \frac{85\pi}{8}$. The answer is $\boxed{\textbf{(C)}\ \frac{85\pi}{8}}$.

The two tangents meeting at $P$ create the classic kite $OAPB$: $OP$ is the perpendicular bisector of the chord $AB$, and triangle $OAP$ is right-angled at $A$. Equal tangent lengths pin down $P$ with one equation, and the altitude-to-hypotenuse relation converts the two known lengths $AM$ and $PM$ into the radius with no messy intersection of lines. Recognize the kite and the right triangle before reaching for coordinates.

Coordinates: $O$ lies on the perpendicular bisector $y - 12 = 3(x - 9)$ and on the line through $A$ perpendicular to $PA$ (slope $-\frac{1}{13}$). Solving gives $O = \left(\frac{37}{4}, \frac{51}{4}\right)$, so $r^2 = \left(\frac{13}{4}\right)^2 + \left(\frac14\right)^2 = \frac{170}{16} = \frac{85}{8}$.

• Trying to find the center by intersecting the perpendicular bisector with a guessed tangent line, or forgetting that PA = PB forces P onto the perpendicular bisector of AB.
• Using $AM^2 = PM \cdot PO$ (the wrong altitude relation) instead of $AM^2 = PM \cdot MO$, or reporting $r^2 = \frac{85}{8}$ as the area without $\pi$.