# 1952 AHSME Problems/Problem 8

## Problem

Two equal circles in the same plane cannot have the following number of common tangents. $\textbf{(A) \ }1 \qquad \textbf{(B) \ }2 \qquad \textbf{(C) \ }3 \qquad \textbf{(D) \ }4 \qquad \textbf{(E) \ }\text{none of these}$

## Solution

Two congruent coplanar circles will either be tangent to one another (resulting in $3$ common tangents), intersect one another (resulting in $2$ common tangents), or be separate from one another (resulting in $4$ common tangents). Having only $\boxed{\textbf{(A)}\ 1}$ common tangent is impossible, unless the circles are non-congruent and internally tangent. $[asy] pair A=(1,0), B=(-1,0), C=(-5,0), D=(-6,0), E=(5,0), F=(8,0); draw(circle(A,1)); draw(circle(B,1)); draw(circle(C,1)); draw(circle(D,1)); draw(circle(E,1)); draw(circle(F,1)); draw((0,1)--(0,-1),red); draw((-2,1)--(2,1),red); draw((-2,-1)--(2,-1),red); draw((-7,1)--(-4,1),red); draw((-7,-1)--(-4,-1),red); draw((4,1)--(9,1),red); draw((4,-1)--(9,-1),red); draw((5+0.5sqrt(2),0.5sqrt(2))--(8-0.5sqrt(2),-0.5sqrt(2)),red); draw((5+0.5sqrt(2),-0.5sqrt(2))--(8-0.5sqrt(2),0.5sqrt(2)),red); label("3",(-1.5,0),E); label("1",(0,0.75),N); label("2",(0,-0.75),S); label("1",(-5.5,0.75),N); label("2",(-5.5,-0.75),S); label("1",(6.5,0.75),N); label("2",(6.5,-0.75),S); label("3",(6.5,-0.25),N); label("4",(6.5,0.25),S); [/asy]$

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