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Problems(1)

Third circle radius

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12/4/2005
A circle centered at A A with a radius of 1 and a circle centered at B B with a radius of 4 are externally tangent. A third circle is tangent to the first two and to one of their common external tangents as shown. The radius of the third circle is [asy] size(220); real r1 = 1; real r2 = 3; real r = (r1*r2)/((sqrt(r1)+sqrt(r2))**2); pair A=(0,r1), B=(2*sqrt(r1*r2),r2); dot(A); dot(B); draw( circle(A,r1) ); draw( circle(B,r2) ); draw( (-1.5,0)--(7.5,0) ); draw( A -- (A+dir(210)*r1) ); label("11", A -- (A+dir(210)*r1), N ); draw( B -- (B+r2*dir(330)) ); label("44", B -- (B+r2*dir(330)), N ); label("AA",A,dir(330)); label("BB",B, dir(140));
draw( circle( (2*sqrt(r1*r),r), r )); [/asy] <spanclass=latexbold>(A)</span> 13<spanclass=latexbold>(B)</span> 25<spanclass=latexbold>(C)</span> 512<spanclass=latexbold>(D)</span> 49<spanclass=latexbold>(E)</span> 12 \displaystyle <span class='latex-bold'>(A)</span> \ \frac {1}{3} \qquad <span class='latex-bold'>(B)</span> \ \frac {2}{5} \qquad <span class='latex-bold'>(C)</span> \ \frac {5}{12} \qquad <span class='latex-bold'>(D)</span> \ \frac {4}{9} \qquad <span class='latex-bold'>(E)</span> \ \frac {1}{2}
Pythagorean Theoremgeometry