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, supposing that the point of intersection is $(x, 0)$ , the Pythagorean Theorem gives $\sqrt{(x-6)^2 + 13^2} = \sqrt{(x-12)^2 + 11^2}$ . This simplifies to $x = 5$ .

Further, notice (due to the right angles formed by a radius and its tangent line) that the quadrilateral (a kite) defined by the circle's center, $A$ , $B$ , and $(5, 0)$ is cyclic. Therefore, we can apply Ptolemy's Theorem to give $2\sqrt{170}x = d \sqrt{40}$ , where $x$ is the radius of the circle and $d$ is the distance between the circle's center and $(5, 0)$ . Therefore, $d = \sqrt{17}x$ . Using the Pythagorean Theorem on the triangle formed by the point $(5, 0)$ , either one of $A$ or $B$ , and the circle's center, we find that $170 + x^2 = 17x^2$ , so $x^2 = \frac{85}{8}$ , and thus the answer is $\boxed{\textbf{(C) }\frac{85}{8}\pi}$ .

Solution 2 (coordinate bash)

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

Line $AC$ is $y=13x-65$ . Therefore, the slope of the line perpendicular to $AC$ is $-\frac{1}{13}$ , so its equation is $y=-\frac{x}{13}+\frac{175}{13}$ .

But notice that this line must pass through $A(6, 13)$ and $(x, 3x-15)$ . Hence $3x-15=-\frac{x}{13}+\frac{175}{13} \Rightarrow x=\frac{37}{4}$ . So the center of the circle is $\left(\frac{37}{4}, \frac{51}{4}\right)$ .

Finally, the distance between the center, $\left(\frac{37}{4}, \frac{51}{4}\right)$ , and point $A$ is $\frac{\sqrt{170}}{4}$ . Thus the area of the circle is $\boxed{\textbf{(C) }\frac{85}{8}\pi}$ .

Solution 3

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

Now, we compute $AC=\sqrt{(6-5)^2+(13-0)^2}=\sqrt{170}$ , $AD=\sqrt{(6-9)^2+(13-12)^2}=\sqrt{10}$ , and $DC=\sqrt{(9-5)^2+(12-0)^2}=\sqrt{160}$ .

Therefore $OA = \frac{AC\cdot AD}{DC}=\sqrt{\frac{85}{8}}$ , and consequently, the area of the circle is $\pi\cdot OA^2 = \boxed{\textbf{(C) }\frac{85}{8}\pi}$ .

Solution 4 (how fast can you multiply two-digit numbers?)

Let $(x,0)$ be the intersection on the x-axis. By Power of a Point Theorem, $(x-6)^2+13^2=(x-12)^2+11^2\implies x=5$ . Then the equations are $13(x-6)+13=y$ and $\frac{11}{7}(x-12)+11=y$ for the tangent lines passing $A$ and $B$ respectively. Then the lines normal to them are $-\frac{1}{13}(x-6)+13=y$ and $-\frac{7}{11}(x-12)+11=y$ . Thus,

$-\frac{7}{11}(x-12)+11=-\frac{1}{13}(x-6)+13$ $\frac{13\cdot7x-11x}{13\cdot11}=\frac{84\cdot13-6\cdot11-2\cdot11\cdot13}{11\cdot13}$ $13\cdot7x-11x=84\cdot13-6\cdot11-2\cdot11\cdot13$

After condensing, $x=\frac{37}{4}$ . Then, the center of $\omega$ is $\left(\frac{37}{4}, \frac{51}{4}\right)$ . Apply distance formula. WLOG, assume you use $A$ . Then, the area of $\omega$ is $\sqrt{\frac{1^2}{4^2}+\frac{13^2}{4^2}}^2\pi=\frac{170\pi}{16} \implies \boxed{\textbf{(C) }\frac{85}{8}\pi}.$

Solution 5 (power of a point)

Firstly, the point of intersection of the two tangent lines has an equal distance to points $A$ and $B$ due to power of a point theorem. This means we can easily find the point, which is $(5, 0)$ . Label this point $X$ . $\triangle{XAB}$ is an isosceles triangle with lengths, $\sqrt{170}$ , $\sqrt{170}$ , and $2\sqrt{10}$ . Label the midpoint of segment $AB$ as $M$ . The height of this triangle, or $\overline{XM}$ , is $4\sqrt{10}$ . Since $\overline{XM}$ bisects $\overline{AB}$ , $\overleftrightarrow{XM}$ contains the diameter of circle $\omega$ . Let the two points on circle $\omega$ where $\overleftrightarrow{XM}$ intersects be $P$ and $Q$ with $\overline{XP}$ being the shorter of the two. Now let $\overline{MP}$ be $x$ and $\overline{MQ}$ be $y$ . By Power of a Point on $\overline{PQ}$ and $\overline{AB}$ , $xy = (\sqrt{10})^2 = 10$ . Applying Power of a Point again on $\overline{XQ}$ and $\overline{XA}$ , $(4\sqrt{10}-x)(4\sqrt{10}+y)=(\sqrt{170})^2=170$ . Expanding while using the fact that $xy = 10$ , $y=x+\frac{\sqrt{10}}{2}$ . Plugging this into $xy=10$ , $2x^2+\sqrt{10}x-20=0$ . Using the quadratic formula, $x = \frac{\sqrt{170}-\sqrt{10}}{4}$ , and since $x+y=2x+\frac{\sqrt{10}}{2}$ , $x+y=\frac{\sqrt{170}}{2}$ . Since this is the diameter, the radius of circle $\omega$ is $\frac{\sqrt{170}}{4}$ , and so the area of circle $\omega$ is $\frac{170}{16}\pi = \boxed{\textbf{(C) }\frac{85}{8}\pi}$ .

Video Solution

For those who want a video solution: (Is similar to Solution 1) https://youtu.be/WI2NVuIp1Ik

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

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