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Difference between revisions of "2009 AMC 12A Problems/Problem 16"
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Simplifying, we obtain <math>r^2 - 8r + 8 = 0</math>. By [[Vieta's formulas]] the sum of the two roots of this equation is <math>\boxed{8}</math>.
 
Simplifying, we obtain <math>r^2 - 8r + 8 = 0</math>. By [[Vieta's formulas]] the sum of the two roots of this equation is <math>\boxed{8}</math>.
  
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(We should actually solve for <math>r</math> to verify that there are two distinct positive roots. In this case we get <math>r=4\pm 2\sqrt 2</math>. This is generally a good rule of thumb, but is not necessary as all of the available answers are integers, and the equation obviously doesn't factor as integers.)
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(We should actually solve for <math>r</math> to verify that there are two distinct positive roots. In this case we get <math>r=4\pm 2\sqrt 2</math>. This is generally a good rule of thumb, but is not necessary as all of the available answers are integers, and the equation obviously doesn't factor as integers. You can also tell that there are two positive roots based on the visual interpretation.)
  
 
<asy>
 
<asy>
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{{AMC12 box|year=2009|ab=A|num-b=15|num-a=17}}
 
{{AMC12 box|year=2009|ab=A|num-b=15|num-a=17}}
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{{MAA Notice}}

Latest revision as of 20:07, 16 January 2020

Problem

A circle with center $C$ is tangent to the positive $x$ and $y$-axes and externally tangent to the circle centered at $(3,0)$ with radius $1$. What is the sum of all possible radii of the circle with center $C$?

$\textbf{(A)}\ 3 \qquad \textbf{(B)}\ 4 \qquad \textbf{(C)}\ 6 \qquad \textbf{(D)}\ 8 \qquad \textbf{(E)}\ 9$

Solution

Let $r$ be the radius of our circle. For it to be tangent to the positive $x$ and $y$ axes, we must have $C=(r,r)$. For the circle to be externally tangent to the circle centered at $(3,0)$ with radius $1$, the distance between $C$ and $(3,0)$ must be exactly $r+1$.

By the Pythagorean theorem the distance between $(r,r)$ and $(3,0)$ is $\sqrt{ (r-3)^2 + r^2 }$, hence we get the equation $(r-3)^2 + r^2 = (r+1)^2$.

Simplifying, we obtain $r^2 - 8r + 8 = 0$. By Vieta's formulas the sum of the two roots of this equation is $\boxed{8}$.

(We should actually solve for $r$ to verify that there are two distinct positive roots. In this case we get $r=4\pm 2\sqrt 2$. This is generally a good rule of thumb, but is not necessary as all of the available answers are integers, and the equation obviously doesn't factor as integers. You can also tell that there are two positive roots based on the visual interpretation.)

[asy] unitsize(0.5cm); defaultpen(0.8); filldraw( Circle( (3,0), 1 ), lightgray, black ); draw( (0,0) -- (15,0), Arrow ); draw( (0,0) -- (0,15), Arrow ); draw( (0,0) -- (15,15), dashed ); real r1 = 4 - 2*sqrt(2), r2 = 4 + 2*sqrt(2); pair S1=(r1,r1), S2=(r2,r2); dot(S1); dot(S2); dot((3,0)); draw( Circle(S1,r1) ); draw( Circle(S2,r2) ); [/asy]

See Also

2009 AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 15 		Followed by
Problem 17
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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