# 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.

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