I have drawn the figure also and plz solve by taking side AP and Bp as radius fast plz....answer in most easy way and in my way....Plz no Google ans and no copy paste...???
Dear Student,
O and O' are the centres of circles having radius 3 cm and 4 cm respectively. OP and OP' are tangents to the given circles. The given circles intersects in P and Q, Suppose OO' intersect PQ in R.
In ΔOPO' and ΔOQO',
OP = OQ (Radius of circle having centre O)
OO' = OO' (Common)
O'P = O'Q (Radius of circle having centre O')
∴ ΔOPO' 
ΔOQO' (SSS congruence criterion)
⇒ ∠POO' = ∠QOO' (CPCT)
In ΔOPR and ΔOQR,
OP = OQ (Radius of circle having centre O)
∠POR = ∠QOR (Proved)
OR = OR (Common)
∴ ΔOPR ΔOQR (SAS congruence criterion)
⇒ ∠ORP = ∠ORQ (CPCT)
∠ORP + ∠ORQ = 180° (Linear pair)
∴ 2∠ORP = 180°
⇒ ∠ORP = 90°
∴ PR = RQ (Perpendicular from the centre of the circle to the chord, bisect the chord)
OP ⊥ O'P (Radius is perpendicular to the tangent at point of contact)
In ΔOPO'
(OO')2 = (OP)2 + (O'P)2
∴ (OO')2 = (3 cm)2 + (4 cm)2 = 25 cm2
⇒ OO' = 5 cm
Let OR = x
∴ O'R = 5 – x
In ΔOPR,
OR2 + PR2 = OP2
∴ x2 + PR2 = (3)2 = 9
⇒ PR2 = 9 – x2 ...(1)
In ΔO'PR,
O'R2 + PR2 = O'P2
∴ (5 – x)2 + PR2 = (4)2 = 16
⇒ 25 + x2 – 10x + 9 – x2 = 16 [Using (1)]
⇒ 34 – 10x = 16
⇒ 10x = 34 – 16 = 18
Thus, the length of the common chord PQ is 
= 4.8 cm.
Regards,