It is given that
`PA=5 cm ,PB=3cm` and `AB=4cm`.
`rArrPA^2=(5)^2=25`.
`PB^2+AB^2=(3)^2+(4)^2=25`
`therefore PA^2=PB^2+AB^2`
`therefore` Angle opposite to side PA must be of `90^@` (converse of phythagoras theorem)
`thereforeanglePBA=90^@`
It is possible only when B coincides with M i.e., centre of smaller circle will lie on M.
(`because` the line segment joining the centres of two circles bisect the common chord at right angle)
So, correct figure must be as shown. This figure satisfies all the conditions
`therefore PQ=2PB` (`because` perpendicular drawn from the centre to the chord bisects the chord)
`=2xx3=6cm`.
