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Difference between revisions of "2009 AMC 12A Problems/Problem 16"

(New page: == Problem == A circle with center <math>C</math> is tangent to the positive <math>x</math> and <math>y</math>-axes and externally tangent to the circle centered at <math>(3,0)</math> with...)
 
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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>.
  
−
(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>.)
+
(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.)
  
 
<asy>
 
<asy>

Revision as of 12:59, 5 August 2012

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.)

[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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