Question

From an external point P, a tangent PT and a line segment PAB is drawn to circle with center O, ON is perpendicular to the chord AB . Prove that PA × PB=PN2-AN2

Answer 1

Given PAB is the secant and PT is the tangent to the circle with centre ODraw ON ⊥ AB
Therefore, AN = BNAlso we have PT2= PA × PB [By tangent secant property] ---- 1
Consider ΔONAOA2= ON2+ AN2 ---- (2)
Similarly inΔPTOOP2= OT2+ PT2
That is PT2= OP2- OT2= ON2+ PN2- OA2 [Since OA = OT (radii)]= ON2+ PN2– ON2– AN2⇒ PT2= PN2– AN2----(3)
From (1) and (3), we get
PA × PB = PN2– AN2.

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