Difference between revisions of "2006 AMC 12A Problems/Problem 19"

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== Problem ==
 
== Problem ==
 
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[[Circle]]s with [[center]]s <math>(2,4)</math> and <math>(14,9)</math> have [[radius | radii]] <math>4</math> and <math>9</math>, respectively. The equation of a common external [[tangent line|tangent]] to the circles can be written in the form <math>y=mx+b</math> with <math>m>0</math>. What is <math>b</math>?
 
[[Circle]]s with [[center]]s <math>(2,4)</math> and <math>(14,9)</math> have [[radius | radii]] <math>4</math> and <math>9</math>, respectively. The equation of a common external [[tangent line|tangent]] to the circles can be written in the form <math>y=mx+b</math> with <math>m>0</math>. What is <math>b</math>?
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<center>[[Image:AMC12_2006A_19.png]]</center>
  
 
<math> \mathrm{(A) \ } \frac{908}{199}\qquad \mathrm{(B) \ } \frac{909}{119}\qquad \mathrm{(C) \ } \frac{130}{17}\qquad \mathrm{(D) \ } \frac{911}{119}</math><math>\mathrm{(E) \ }  \frac{912}{119}</math>
 
<math> \mathrm{(A) \ } \frac{908}{199}\qquad \mathrm{(B) \ } \frac{909}{119}\qquad \mathrm{(C) \ } \frac{130}{17}\qquad \mathrm{(D) \ } \frac{911}{119}</math><math>\mathrm{(E) \ }  \frac{912}{119}</math>

Revision as of 19:29, 8 November 2007

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}{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

This solution needs a clearer explanation and a diagram.

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,

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

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 (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