Answer to Question #13555 in Geometry for john
r^2 = 9 pi / pi;
r = 3;
So, AB = 3;
BC^2 = AC^2 - AB^2 = 25 - 9 = 16;
BC = 4;
Let O be the point of tangency of the circles with the centers B and C respectively.
BC = BO + OC;
BO = r = 3 => OC = 1;
P_big = 2*pi*BO = 6*pi;
P_small = 2*pi*C0 = 2*pi
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