2019 AMC 10B Problems/Problem 23

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 1

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 $\boxed{\textbf{(C) }\frac{85}{8}\pi}$ .

Solution 2

First, follow solution 1 and obtain $x=5$ . Label the point $(5,0)$ as point $C$ . The midpoint $M$ of segment $AB$ is $(9, 12)$ . Notice that the center of the circle must lie on the line that goes through the points $C$ and $M$ . Thus, the center of the circle lies on the line $y=3x-15$ .

Line $AC$ is $y=13x-65$ . The perpendicular line must pass through $A(6, 13)$ and $(x, 3x-15)$ . The slope of the perpendicular line is $-\frac{1}{13}$ . The line is hence $y=-\frac{x}{13}+\frac{175}{13}$ . The point $(x, 3x-15)$ lies on this line. Therefore, $3x-15=-\frac{x}{13}+\frac{175}{13}$ . Solving this equation tells us that $x=\frac{37}{4}$ . So the center of the circle is $(\frac{37}{4}, \frac{51}{4})$ . The distance between the center, $(\frac{37}{4}, \frac{51}{4})$ , and point A is $\frac{\sqrt{170}}{4}$ . Hence, the area is $\frac{85}{8}\pi$ . The answer is $\boxed{\textbf{(C) }\frac{85}{8}\pi}$ .

Solution 3

The mid point of $AB$ is $D(9,12)$ . Let the tangent lines at $A$ and $B$ intersect at $C(a,0)$ on the $X$ axis. Then $CD$ would be the perpendicular bisector of $AB$ . Let the center of circle be O. Then $\triangle AOC$ is similar to $\triangle DAC$ , that is $\frac{OA}{AC} = \frac{AD}{DC}.$ The slope of $AB$ is $\frac{13-11}{6-12}=\frac{-1}{3}$ , therefore the slope of CD will be 3. The equation of $CD$ is $y-12=3*(x-9)$ , that is $y=3x-15$ . Let $y=0$ . Then we have $x=5$ , which is the $x$ coordinate of $C(5,0)$ .

$AC=\sqrt{(6-5)^2+(13-0)^2}=\sqrt{170}$ $AD=\sqrt{(6-9)^2)+(13-12)^2}=\sqrt{10}$ $DC=\sqrt{(9-5)^2+(12-0)^2}=\sqrt{160}$ Therefore $OA = \frac{AC*AD}{DC}=\sqrt{\frac{85}{5}}$ Consequently, the area of the circle is $pi* OA^2 = pi*\frac{85}{5}$ (by Zhen Qin) (P.S. Will someone please Latex this?) ( $\LaTeX$ ed by a pewdiepie subscriber)

2019 AMC 10B (Problems • Answer Key • Resources)
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 (Problems • Answer Key • Resources)
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

Page

Toolbox

Search

2019 AMC 10B Problems/Problem 23

Contents

Problem

Solution 1

Solution 2

Solution 3

See Also