Question

Points $A=(6,13)$ and $B=(12,11)$ lie on circle $\omega$ in the plane. Suppose that the tangent lines to $\omega$ at $A$ and $B$ intersect at a point on the $x$ -axis. What is the area of $\omega?$

$\dfrac{83\pi}{8}$

$\dfrac{21\pi}{2}$

$\dfrac{85\pi}{8}$

$\dfrac{43\pi}{4}$

$\dfrac{87\pi}{8}$

Solution:

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.

2019 AMC 10B Problem 23

Solution:

Problem 23 in Other Years