2006 AMC 12A Problems/Problem 19

Revision as of 16:51, 28 September 2014 by Mathgeek2006 (talk | contribs) (Problem)

Problem

Circles with centers $(2,4)$ and $(14,9)$ have radii $4$ and $9$, respectively. The equation of a common external tangent to the circles can be written in the form $y=mx+b$ with $m>0$. What is $b$?

AMC12 2006A 19.png

$\mathrm{(A) \ } \frac{908}{119}\qquad \mathrm{(B) \ } \frac{909}{119}\qquad \mathrm{(C) \ } \frac{130}{17}\qquad \mathrm{(D) \ } \frac{911}{119}\qquad \mathrm{(E) \ }  \frac{912}{119}$

Solution

Let $L_1$ be the line that goes through $(2,4)$ and $(14,9)$, and let $L_2$ be the line $y=mx+b$. If we let $\theta$ be the measure of the acute angle formed by $L_1$ and the x-axis, then $\tan\theta=\frac{5}{12}$. $L_1$ clearly bisects the angle formed by $L_2$ and the x-axis, so $m=\tan{2\theta}=\frac{2\tan\theta}{1-\tan^2{\theta}}=\frac{120}{119}$. We also know that $L_1$ and $L_2$ intersect at a point on the x-axis. The equation of $L_1$ is $y=\frac{5}{12}x+\frac{19}{6}$, so the coordinate of this point is $\left(-\frac{38}{5},0\right)$. Hence the equation of $L_2$ is $y=\frac{120}{119}x+\frac{912}{119}$, so $b=\frac{912}{119}$, and our answer choice is $\boxed{\mathrm{E}}$.

See also

2006 AMC 12A (ProblemsAnswer KeyResources)
Preceded by
Problem 18
Followed by
Problem 20
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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