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Find [K]:[M]

Source: 1976 AHSME Problem 24

May 19, 2014
ratiogeometrygeometric transformationhomothetyPythagorean TheoremAMC

Problem Statement

[asy] size(150); pair A=(0,0),B=(1,0),C=(0,1),D=(-1,0),E=(0,.5),F=(sqrt(2)/2,.25); draw(circle(A,1)^^D--B); draw(circle(E,.5)^^circle( F ,.25)); label("AA", D, W); label("KK", A, S); label("BB", B, dir(0)); label("LL", E, N); label("MM",shift(-.05,.05)*F); //Credit to Klaus-Anton for the diagram[/asy]
In the adjoining figure, circle K\mathit{K} has diameter AB\mathit{AB}; cirlce L\mathit{L} is tangent to circle K\mathit{K} and to AB\mathit{AB} at the center of circle K\mathit{K}; and circle M\mathit{M} tangent to circle K\mathit{K}, to circle L\mathit{L} and AB\mathit{AB}. The ratio of the area of circle K\mathit{K} to the area of circle M\mathit{M} is
<spanclass=latexbold>(A)</span>12<spanclass=latexbold>(B)</span>14<spanclass=latexbold>(C)</span>16<spanclass=latexbold>(D)</span>18<spanclass=latexbold>(E)</span>not an integer<span class='latex-bold'>(A) </span>12\qquad<span class='latex-bold'>(B) </span>14\qquad<span class='latex-bold'>(C) </span>16\qquad<span class='latex-bold'>(D) </span>18\qquad <span class='latex-bold'>(E) </span>\text{not an integer}