2006 AMC 12A Problem 19

Attempt Problem 19 of the 2006 AMC 12A below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2006 AMC 12A solutions, or check the answer key.

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19.

Circles with centers (2,4)(2, 4) and (14,9)(14, 9) have radii 44 and 9,9, respectively. The equation of a common external tangent to the circles can be written in the form y=mx+by = mx + b with m>0.m \gt 0. What is b?b?

908119\dfrac{908}{119}

909119\dfrac{909}{119}

13017\dfrac{130}{17}

911119\dfrac{911}{119}

912119\dfrac{912}{119}

Answer: E
Concepts:tangent linetrigonometric identitycoordinate geometry
Difficulty rating: 1960
Solution:

Each circle's radius equals its center's yy-coordinate, so both are tangent to the xx-axis, which is a common external tangent. The two external tangents meet at the xx-intercept of the line through the centers.

That line has slope 94142=512=tanθ\tfrac{9 - 4}{14 - 2} = \tfrac{5}{12} = \tan\theta and passes through (2,4),(2, 4), meeting the xx-axis at (385,0).\left(-\tfrac{38}{5}, 0\right).

The other tangent makes angle 2θ2\theta with the xx-axis, so its slope is tan2θ=25121(512)2=120119. \tan 2\theta = \frac{2 \cdot \tfrac{5}{12}}{1 - \left(\tfrac{5}{12}\right)^2} = \frac{120}{119}. Then b=120119385=912119.b = \tfrac{120}{119} \cdot \tfrac{38}{5} = \tfrac{912}{119}.

Thus, the correct answer is E.

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