Problem:
Two circles are externally tangent. Lines PAB and PA′B′ are common tangents with A and A′ on the smaller circle and B and B′ on the larger circle. If PA=AB=4, then the area of the smaller circle is
Answer Choices:
A. 1.44Ï€
B. 2Ï€
C. 2.56Ï€
D. 8​π
E. 4Ï€
Solution:
Let C be the center of the smaller circle, T be the point where the two circles are tangent, and X be the intersection of the common internal tangent with AB. Since tangents from a common point are equal, BX=TX=AX=2AB​=2. Since △ACP∼△TXP, it follows that
TXAC​=TPAP​, or 2AC​=62−22​4​, so AC=2​.
Hence the area of the circle with radius AC is 2Ï€.
OR
Let C1​,C2​,r and R be the centers and radii of the smaller and larger circle, respectively. Points P,C1​ and C2​ are collinear by symmetry. Since the right triangles PAC1​ and PBC2​ are similar,
Rr​=PC2​PC1​​=PBPA​=84​
Thus R=2r and PC1​=C1​C2​=R+r=3r. Apply the Pythagorean theorem to △PAC1​ to find 42+r2=(3r)2, r2=2 and πr2=2π.
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