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.


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) $AOBX$ is cyclic.

Therefore, we can apply Ptolemy's Theorem to give:

$2\sqrt{170}r = d \sqrt{40}$, where $r$ is the radius of the circle and $d$ is the distance between the circle's center and $(5, 0)$. Therefore, $d = \sqrt{17}r$.

Using the Pythagorean Theorem on the right triangle $OAX$ (or $OBX$), we find that $170 + r^2 = 17r^2$, so $r^2 = \frac{85}{8}$, and thus the area of the circle is $\boxed{\textbf{(C) }\frac{85}{8}\pi}$.

Diagram for Solution 1



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 for the tangent lines passing $A$ and $B$, respectively, are $13(x-6)+13=y$ and $\frac{11}{7}(x-12)+11=y$. Then the lines normal (perpendicular) to them are $-\frac{1}{13}(x-6)+13=y$ and $-\frac{7}{11}(x-12)+11=y$. Solving for $x$, we have

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

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 \[\left(\sqrt{\frac{1^2}{4^2}+\frac{13^2}{4^2}}\right)^2\pi=\frac{170\pi}{16} \implies \boxed{\textbf{(C) }\frac{85}{8}\pi}.\]

Solution 5 (tangent cheese)

After getting $x=5$, let $C=(5,0)$. Get the slopes of the lines $AC$ and $BC$, namely $\frac{13}{6-5}=13$, $\frac{11}{12-5}=\frac{11}{7}$. Then, use tangent angle subtraction to get $\tan{2x}=\frac{13-\frac{11}{7}}{1+13*\frac{11}{7}}=\frac{80}{150}=\frac{8}{15}$. Then, apply tangent double angle to get $\tan{2x}=\frac{8}{15}=\frac{2\tan{x}}{1-\tan^2{x}}$. Solving, we obtain $\tan{x}=\frac{1}{4}$. Then, note that $\tan{x}=r/{BC}$, so $r=\frac{1}{4}*\sqrt{170}$. Finishing off, we obtain $A=\pi*r^2=\pi*170/16=\boxed{\textbf{(C) }\frac{85}{8}\pi}$.


Video Solution

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

Video Solution by TheBeautyofMath



Video Solution by The Power of Logic


~The Power of Logic

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

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