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 $\boxed{\textbf{(C) }\frac{85}{8}\pi}$ . -heon01px2020 (edited by molocyxu)

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

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

The mid point of AB is D(9,12), suppose the tanget lines at A and B intersect at C(a,0)on X axis, CD would be the perpendicular bisector of AB. Suppose the center of circle is O, then triangle AOC similiar to DAC, that is OA/AC=AD/DC. The slope of AB is (13-11)/(6-12)=-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, e have x=5, which is the x coordiante 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)=aqrt(160) Therefore OA=AC*AD/DC=sqrt(85/5) area of the circle is pi*OA^2=pi*85/5

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2019 AMC 10B Problems/Problem 23

Contents

Problem

Solution 1

Solution 2

See Also