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 $P=(x,0)$ be the intersection. Equal tangent lengths give $PA=PB,$ so $(x-6)^2+13^2=(x-12)^2+11^2,$ yielding $x=5$ and $P=(5,0).$

The center $O=(h,k)$ satisfies $OA\perp PA$ and $OB\perp PB.$ With $PA=(1,13)$ and $PB=(7,11),$ these give $h+13k=175$ and $7h+11k=205,$ so $O=\left(\dfrac{37}{4},\dfrac{51}{4}\right).$

Then $r^2=OA^2=\left(\dfrac{13}{4}\right)^2+\left(\dfrac14\right)^2=\dfrac{170}{16}=\dfrac{85}{8},$ so the area is $\dfrac{85\pi}{8}.$

Thus, C is the correct answer.

2019 AMC 12B Problem 20

Solution:

Problem 20 in Other Years