two circles with centres O and O' of radii 3cm and 4cm respectively intersect at two points P and Q such that OP and O'P are tangents to the two circles find the lenghth of the common chord PQ.
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.