Question

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 $m \gt 0.$ What is $b?$

$\dfrac{908}{119}$

$\dfrac{909}{119}$

$\dfrac{130}{17}$

$\dfrac{911}{119}$

$\dfrac{912}{119}$

Solution:

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

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

The other tangent makes angle $2\theta$ with the $x$ -axis, so its slope is $\tan 2\theta = \frac{2 \cdot \tfrac{5}{12}}{1 - \left(\tfrac{5}{12}\right)^2} = \frac{120}{119}.$ Then $b = \tfrac{120}{119} \cdot \tfrac{38}{5} = \tfrac{912}{119}.$

Thus, the correct answer is E.

2006 AMC 12A Problem 19

Solution:

Problem 19 in Other Years