AB is the diameter of a circle with radius 5. Q is on AB, and P is on the circle, so that PQ is perpendicular to AB. If AQ = 2, then find PB.
+47 \(A_Q ={1\over5}AB\)
\(Arc[AP]= {1\over 5}Arc[AB]\)
\(PQ = { \sqrt{(P_O)^2-(AB)^2}}\)
\(PQ = 4\)
\(PB = {\sqrt{(PQ)^2+(QB)^2} }\)
\(PB =\sqrt{80}\)
R:
\(PB = {4\sqrt{5}}\)
