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2006 AMC 12A Problems/Problem 19

Revision as of 19:51, 2 February 2007 by Azjps (talk | contribs) (→‎Solution: solution)

Problem

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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 $b$?

$\mathrm{(A) \ } \frac{908}{199}\qquad \mathrm{(B) \ } \frac{909}{119}\qquad \mathrm{(C) \ } \frac{130}{17}\qquad \mathrm{(D) \ } \frac{911}{119}$$\mathrm{(E) \ } \frac{912}{119}$

Solution

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Notice that both circles are tangent to the x-axis and each other. Call the circles (respectively) A and B; the distance between the two centers is $4 + 9 = 13$. If we draw the parallel radii that lead to the common external tangent, a line can be extended parallel to the tangent from A to the radius of circle B. This creates a 5-12-13 triangle. To find the slope of that line (which is parallel to the tangent), note that another 5-12-13 triangle can be drawn below the first one such that the side with length 12 is parallel to the x-axis. The slope can be found by using the double tangent identity,

$\tan (2 \tan ^{-1} (\frac{5}{12}) = \frac{\frac{5}{12} + \frac{5}{12}}{1 - \frac{5}{12}\frac{5}{12}}$
$= \frac{120}{119}$

To find the x and y coordinates of the point of tangency of circle A, we can set up a ratio (the slope will be –119/120 because it is the negative reciprocal):

$\frac{119}{\sqrt{119^2 + 120^2}}$ $=$ $\frac{119}{169} = \frac{y - 4}{4}$
$\frac{-120}{\sqrt{119^2 + 120^2}} = \frac{-120}{169} = \frac{x - 2}{4}$
$x = \frac{-142}{169}, y = \frac{1152}{169}$

We can plug this into the equation of the line for the tangent to get:

$\frac{1152}{169} = \frac{120}{119}\frac{-142}{169} + b$
$b = \frac{912}{119}$ $\Rightarrow \mathrm{E}$

See also

2006 AMC 12A (Problems • Answer Key • Resources)
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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