A and B are the centres of the circles with radii 5cm and 3cm respectively.
C is the mid-point of AB.
Extend AB upto O point on circumference of outer circle.
AB=AO−BO=5−3=2cm (since AO and BO are radii of larger and smaller circles)
AC=AB2=22=1cm
now in right angled triangle AMP
AC=1cm,AP=5cm
by pythagoras thm.
AP2=PC2+AC2
PC2=√AP2−AC2
PC2=√52−12
Therefore PQ=2PC=2.√24=4.√6cm [CP=CQ]
So , option A is the answer.