2019 AMC 10B Problems/Problem 23

Revision as of 17:07, 14 February 2019 by Expilncalc (talk | contribs) (Solution: Added solution.)
The following problem is from both the 2019 AMC 10B #23 and 2019 AMC 12B #20, so both problems redirect to this page.

Problem

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 $\omega$?

$\textbf{(A) }\frac{83\pi}{8}\qquad\textbf{(B) }\frac{21\pi}{2}\qquad\textbf{(C) } \frac{85\pi}{8}\qquad\textbf{(D) }\frac{43\pi}{4}\qquad\textbf{(E) }\frac{87\pi}{8}$

Solution

First, observe that the two tangent lines are of identical length. Therefore, suppose the intersection was $(x, 0)$. Using Pythagorean Theorem gives $x=5$.

Notice (due to the right angles formed by a radius and its tangent line) that the quadrilateral (kite) defined by circle center, $A$, $B$, and $(5, 0)$ form a cyclic quadrilateral. Therefore, we can use Ptolemy's theorem:

$2\sqrt{170}x = d * \sqrt{40}$, where $d$ represents the distance between circle center and $(5, 0)$. Therefore, $d = \sqrt{17}x$. Using Pythagorean Theorem on $(5, 0)$, either one of $A$ or $B$, and the circle center, we realize that $170 + x^2 = 17x^2$, at which point $x^2 = \frac{85}{8}$, so the answer is $\boxed{\textbf{(C) }\frac{85}{8}\pi}$.

See Also

2019 AMC 10B (ProblemsAnswer KeyResources)
Preceded by
Problem 22
Followed by
Problem 24
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 10 Problems and Solutions
2019 AMC 12B (ProblemsAnswer KeyResources)
Preceded by
Problem 19
Followed by
Problem 21
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 12 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png