2019 AMC 10B Problem 23

Question

Points

A=(6,13)

and

B=(12,11)

lie on a 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}

Answer: C

Concepts:coordinate geometry tangent line perpendicular bisector

Difficulty rating: 2150

Small hint:

The tangent intersection lies on the perpendicular bisector of

AB

Big hint:

The radius to a tangent point is perpendicular to the tangent line

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.