Problem:
A circle of radius r has chords AB of length 10 and CD of length 7. When AB and CD are extended through B and C, respectively, they intersect at P, which is outside the circle. If ∠APD=60∘ and BP=8, then r2=
Answer Choices:
A. 70
B. 71
C. 72
D. 73
E. 74
Solution:
Using properties of secant segments (power of a point), it follows that PD⋅PC=PA⋅PB=18⋅8. But PD⋅PC=(PC+7)⋅PC, so PD=16 and PC=9. Since AP=2PC and ∠APC=60∘,∠ACP=90∘. (Why?) It follows that AC=93 and ∠ACD is also a right angle. Thus AD is a diameter of the circle. Apply the Pythagorean Theorem to triangle ACD to obtain
\end
