PAB is a secant and PT is a tangent. Prove that PA X PB =PT2

Two circles intersect each other at A and B. The common chord AB is produced to meet common tangent PQ to the circle at D. Prove that DP = DQ.

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Given: A secant PAB to a circle C(O, r) intersect it in A and B and PT is a tangent.

To prove: PA × PB = PT2

Construction: Draw OD ⊥ AB. Join OP, OT and OA.

Proof:

Since, OD ⊥ AB

∴ AD = DB ...(1) (Perpendicular from the centre to the chord bisects the chord)

PA × PB = (PD – AD) (PD + BD)

= (PD – AD) (PD + AD) [Using (1)]

= PD2 – AD2

In right ΔOPD,

OP2 = OD2 + PD2

⇒ PD2 = OP2 – OD2

∴ PA × PB = (OP2 – OD2) – AD2 = OP2 – (OD2 + AD2)

In right ΔOAD,

OA2 = OD2 + AD2

∴ PA × PB = OP2 – OA2 = OP2 – OT2 (OA = OT)

In ΔOPT,

OP2 = PT2 + OT2

⇒ OP2 – OT= PT2

∴ PA × PB = PT2

 

it was already answered by an expert for someone else

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THUMBS UP!!!!
  • -10

Hence proved
Thumbs up please
Medha
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Yor question Iss wrong
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Lahri dabang

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