Let PT be a tangent to the circle from an external point P and a secant to the circle through P intersects the circle at points A and B, then PT ² = PA � PB”.

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From the figure, each and everything is clear. The proof is as follows: 

Given: A secant PAB to a circle with centre O and radius r, intersects it at 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]

Consider, PA × PB = (PD – AD) (PD + BD)

  = (PD – AD) (PD + AD) [Using (1)]

   = PD² – AD²   ... (2)

In right ΔOPD, by Pythagoras theorem we have

OP² = OD² + PD²

⇒ PD² = OP² – OD²   ... (3)

Using result (3) in (2), we have

PA × PB = (OP² – OD²) – AD²

 = OP² – (OD² + AD²) ... (4)

In ΔOAD, OA² = OD²+ AD²   ... (5)

Using result (5) in (4)

PA × PB = OP² – OA² = OP² – OT²     ...(6) [OA = OT = radius]

In ΔOPT,

OP² = OT² + PT²

⇒ OP² – OT² = PT²   ... (7)

Using result (7) in (6), we have 

PA × PB = PT².

Hence proved.