Problem:
Two circles are externally tangent. Lines PAB and PAβ²Bβ² are common tangents with A and Aβ² on the smaller circle and B and Bβ² on the larger circle. If PA=AB=4, then the area of the smaller circle is
Answer Choices:
A. 1.44Ο
B. 2Ο
C. 2.56Ο
D. 8βΟ
E. 4Ο
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=2ABβ=2. Since β³ACPβΌβ³TXP, it follows that
TXACβ=TPAPβ, or 2ACβ=62β22β4β, so AC=2β.
Hence the area of the circle with radius AC is 2Ο.
OR
Let C1β,C2β,r and R be the centers and radii of the smaller and larger circle, respectively. Points P,C1β and C2β are collinear by symmetry. Since the right triangles PAC1β and PBC2β are similar,
Rrβ=PC2βPC1ββ=PBPAβ=84β
Thus R=2r and PC1β=C1βC2β=R+r=3r. Apply the Pythagorean theorem to β³PAC1β to find 42+r2=(3r)2, r2=2 and Οr2=2Ο.
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