1996 AMC 12 Problem 18

Question

A circle of radius

2

has center at

(2,0).

A circle of radius

1

has center at

(5,0).

A line is tangent to the two circles at points in the first quadrant. Which of the following is closest to the

y

-intercept of the line?

\frac{\sqrt2}{4}

\frac{\sqrt8}{3}

1+\sqrt3

2\sqrt2

3

Write the tangent as

y=mx+b

and use point-to-line distance for each center

Subtract the two distance equations to determine the slope before solving for the intercept

Write the upper common tangent as

mx-y+b=0

and let

s=\sqrt{m^2+1}.

The two center-to-line distances give

2m+b=2s,\qquad 5m+b=s.

Thus

3m=-s,

so

9m^2=m^2+1

and

m=-\frac{1}{\sqrt8}.

From the second equation,

\begin{aligned} b&=s-5m\\ &=\frac3{\sqrt8}+\frac5{\sqrt8}=2\sqrt2. \end{aligned}

Therefore the correct answer is D.

Problem 18 in other years