1996 AHSME Problems/Problem 18

Problem

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?

$\text{(A)}\ \sqrt{2}/4\qquad\text{(B)}\ 8/3\qquad\text{(C)}\ 1+\sqrt 3\qquad\text{(D)}\ 2\sqrt 2\qquad\text{(E)}\ 3$

Solution

The two circles are tangent to eaech other at the point $(4,0)$ , since it is both $2$ units from $(2,0)$ and $1$ unit from $(5,0)$ .

Label the x-intercept of the common tangent line $A$ , and label the y-intercept of the common tangent $B$ . Triangle $\triangle OAB$ is a right triangle at the origin.

Label $D$ the point of tangency to the first, big circle, and label $E$ the point of tangency to the small circle. $\angle D$ and $\angle E$ are both right angles as well.

Label the center of the big circle $F$ , and the small circle $G$ .

$\triangle DAF \sim \triangle EAG \sim \triangle OAB$ because they each have one right angle, and also ahve a common angle $A$ .

The y-intercept is the length $OB$ .

We know that $DF = 2$ and $EG = 1$ because they are radii of circles. From similarity, $\frac{DF}{EG} = \frac{FA}{GA}$ . Thus, $FA = 2GA$ .

Looking at line $FGA$ , we know that $FG + GA = FA$ .

Thus, $FG + GA = 2GA$ , meaning $FG = GA$ . Since $FG = 3$ because it's the distance of the centers of the circles, so $GA = 3$ as well. This gives point $A$ as $(8,0)$ .

Since $\triangle EAG$ is a right triangle with $GA = 3$ and $GE = 1$ , we know that $AE = \sqrt{GA^2 - GE^2} = \sqrt{8}$

Thus, all triangles are $1:\sqrt{8}:3$ triangles.

$\triangle OAC$ has as its middle side $OA$ , which is $8$ . Thus, all sides are scaled up by a factor of $\sqrt{8}$ , and $OB$ , the shortest side, is $\sqrt{8}$ . This means the y-intercept is $\sqrt{8} = 2\sqrt{2}$ , and the answer is $\boxed{D}$ .

1996 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 17	Followed by Problem 19
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1996 AHSME Problems/Problem 18

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