# 2002 AMC 12A Problems/Problem 18

## Problem

Let $C_1$ and $C_2$ be circles defined by $(x-10)^2 + y^2 = 36$ and $(x+15)^2 + y^2 = 81$ respectively. What is the length of the shortest line segment $PQ$ that is tangent to $C_1$ at $P$ and to $C_2$ at $Q$? $\text{(A) }15 \qquad \text{(B) }18 \qquad \text{(C) }20 \qquad \text{(D) }21 \qquad \text{(E) }24$

## Solution 1

(C) First examine the formula $(x-10)^2+y^2=36$, for the circle $C_1$. Its center, $D_1$, is located at (10,0) and it has a radius of $\sqrt{36}$ = 6. The next circle, using the same pattern, has its center, $D_2$, at (-15,0) and has a radius of $\sqrt{81}$ = 9. So we can construct this diagram: $[asy] unitsize(0.3cm); defaultpen(0.8); path C1=circle((10,0),6); path C2=circle((-15,0),9); draw(C1); draw(C2); draw( (-25,0) -- (17,0) ); dot((10,0)); dot((-15,0)); pair[] p1 = intersectionpoints(C1, circle((5,0),5) ); pair[] p2 = intersectionpoints(C2, circle((-7.5,0),7.5) ); dot(p1); dot(p2); draw((10,0)--p1--p2--(-15,0)); label("C_1",(10,0) + 6*dir(-45), SE ); label("C_2",(-15,0) + 9*dir(225), SW ); label("D_1",(10,0), SE ); label("D_2",(-15,0), SW ); label("Q", p2, NE ); label("P", p1, SW ); label("O", (0,0), SW ); [/asy]$ Line PQ is tangent to both circles, so it forms a right angle with the radii (6 and 9). This, as well as the two vertical angles near O, prove triangles S $_2$QO and S $_1$PO similar by AA, with a scale factor of 6:9, or 2:3. Next, we must subdivide the line D $_2$D $_1$ in a 2:3 ratio to get the length of the segments D $_2$O and D $_1$O. The total length is 10 - (-15), or 25, so applying the ratio, D $_2$O = 15 and D $_1$O = 10. These are the hypotenuses of the triangles. We already know the length of D $_2$Q and D $_1$P, 9 and 6 (they're radii). So in order to find PQ, we must find the length of the longer legs of the two triangles and add them. $15^2 - 9^2 = (15-9)(15+9) = 6 \times 24 = 144$ $\sqrt{144} = 12$ $10^2-6^2 = (10-6)(10+6) = 4 \times 16 = 64$ $\sqrt{64} = 8$

Finally, the length of PQ is $12+8=\boxed{20}$, or C.

## Solution 2

Using the above diagram, imagine that segment $\overline{QS_2}$ is shifted to the right to match up with $\overline{PS_1}$. Then shift $\overline{QP}$ downwards to make a right triangle. We know $\overline{S_2S_1} = 25$ from the given information and the newly created leg has length $\overline{QS_2} + \overline{PS_1} = 9 + 6 = 15$. Hence by Pythagorean theorem $15^2 + {\overline{QP}}^2 = 25^2$. $\overline{QP} = \boxed{20}$, or C.

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