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 cm^{2}

⇒ OO' = 5 cm

Let OR = *x*

∴ O'R = 5 – *x*

In ΔOPR,

OR^{2 }+ PR^{2} = OP^{2}

∴ *x*^{2} + PR^{2 }= (3)^{2} = 9

⇒ PR^{2} = 9 – *x*^{2} ...(1)

In ΔO'PR,

O'R^{2 }+ PR^{2} = O'P^{2}

∴ (5 – *x*)^{2 }+ PR^{2} = (4)^{2} = 16

⇒ 25 + *x*^{2} – 10*x* + 9 – *x*^{2} = 16 [Using (1)]

⇒ 34 – 10*x* = 16

⇒ 10*x* = 34 – 16 = 18

Thus, the length of the common chord PQ is = 4.8 cm.

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