2019 AMC 10B Problem 23

Two tangent lines from one outside point give tangent segments of equal length. That fact finds the meeting point quickly.

Step 1: find the meeting point $P$.

1. $P$ is on the $x$-axis, so $P = (p, 0)$. Equal tangent lengths mean $PA = PB$, so $PA^2 = PB^2$:
$$(p - 6)^2 + 13^2 = (p - 12)^2 + 11^2.$$
2. Expand: $p^2 - 12p + 36 + 169 = p^2 - 24p + 144 + 121$, so $12p = 60$ and $p = 5$.
3. So $P = (5, 0)$, and $PA^2 = 1^2 + 13^2 = 170$.

Step 2: the picture.

1. Let $M = (9, 12)$ be the midpoint of $AB$, and let $O$ be the center with radius $r$.
2. $O$ and $P$ are each the same distance from $A$ and from $B$, so both lie on the perpendicular bisector of $AB$. That line passes through $M$. So $P$, $M$, $O$ are on one line, with $M$ between $P$ and $O$.
3. A radius meets its tangent line at a right angle, so triangle $OAP$ has a right angle at $A$.

Step 3: the lengths we know.

1. $AB = \sqrt{6^2 + 2^2} = \sqrt{40}$, so $AM = \frac12\sqrt{40}$ and $AM^2 = 10$.
2. $PM = \sqrt{(9 - 5)^2 + 12^2} = \sqrt{160}$, so $PM = 4\sqrt{10}$.

Step 4: Pythagoras twice. Let $d = MO$, the unknown distance from $M$ to the center.

9. In right triangle $AMO$ (the right angle is at $M$): $r^2 = AM^2 + d^2 = 10 + d^2$.
10. In right triangle $OAP$: $OP^2 = PA^2 + r^2$, where $OP = PM + d = 4\sqrt{10} + d$. So
$$(4\sqrt{10} + d)^2 = 170 + 10 + d^2.$$
11. Expand the left side: $160 + 8\sqrt{10}\,d + d^2 = 180 + d^2$. So $8\sqrt{10}\,d = 20$ and $d = \frac{20}{8\sqrt{10}} = \frac{\sqrt{10}}{4}$, giving $d^2 = \frac{10}{16}$.
12. Then $r^2 = 10 + \frac{10}{16} = \frac{170}{16} = \frac{85}{8}$.

The area is $\pi r^2 = \frac{85\pi}{8}$.

The answer is $\boxed{\textbf{(C)}\ \frac{85\pi}{8}}$.

Two tangents from one point always make a kite shape with the center: the tangent segments are equal, the line from the point to the center cuts the chord $AB$ in half at a right angle, and each radius meets its tangent at a right angle. Once you see those right triangles, the Pythagorean theorem does the rest. Equal tangent lengths are the first thing to try whenever two tangents meet.

Use the altitude rule in a right triangle: in triangle $OAP$ with the right angle at $A$, the altitude $AM$ to the long side satisfies $AM^2 = PM \cdot MO$. So $MO = \frac{10}{4\sqrt{10}} = \frac{\sqrt{10}}{4}$, and $r^2 = AM^2 + MO^2 = \frac{85}{8}$ as before.