
Andrew D. answered 06/05/15
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Pranav,
First of all, let's emend the question to state "AB is a common...
Draw in the sides of quadrilateral ABPQ and the line QM parallel to BA and meeting AP at M.
Angles PAB and ABQ are right angles, so ABQM is a rectangle and therefore AM=3 and MQ=8
Consider the right triangle PMQ and apply Pythagoras' Theorem: PQ^2=MP^2+MQ^2
PQ^2=64+9
PQ=sqrt(73)
You might want to prove to yourself that the circles intersect...Although you drew the diagram initially, perhaps
you did not know whether to draw the circles disjoint, tangent or intersecting?