2006 AMC 12A Problem 19

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}

Each circle is tangent to the

x

-axis, so the

x

-axis is one common external tangent

The two external tangents meet on the line through the centers, whose slope is

\tfrac{5}{12} = \tan\theta;

the other tangent has slope

\tan 2\theta

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.

Problem 19 in other years