2019 AMC 10B Problem 23

Let $P$ be the intersection point of the two tangents. Since tangent lengths from the same point are equal, $PA=PB$, so $P$ lies on the perpendicular bisector of $\overline{AB}$. The midpoint of $A(6,13)$ and $B(12,11)$ is $(9,12)$, and the slope of $AB$ is $-\dfrac13$, so the perpendicular bisector is $y=3x-15$. Its intersection with the $x$-axis is $P=(5,0)$. The tangent line through $P$ and $A$ has slope $13$, so the radius to $A$ has slope $-\dfrac1{13}$. Intersecting $y-13=-\dfrac1{13}(x-6)$ with $y=3x-15$ gives center $\left(\dfrac{37}{4},\dfrac{51}{4}\right)$. Thus $r^2=\left(\dfrac{37}{4}-6\right)^2$ $+\left(\dfrac{51}{4}-13\right)^2$ $=\dfrac{85}{8}$, so the area is $\dfrac{85\pi}{8}$. Thus, C is the correct answer.