Question

in the above given below figure from an external point P ,a tangent PT and a secant PAB is drawn to circle with center O.ON is perpendicular on chord AB.Prove that PA.PB=PT2

Answer 1

Draw ON ⊥AB
Therefore, AN = BN
Also we have PT= PA × PB [By tangent secant property] ---- (1)
Consider ΔONA
OA²= ON²+ AN²---- (2)
Similarly inΔPTO
OP²= OT²+ PT²
That is PT² = OP² - OT²
= ON² + PN² - OA²  [Since OA = OT (radii)]
= ON+ PN²  – ON² – AN²
⇒ PT² = PN²– AN²----(3)
From (1) and (3), we get
PA × PB = PN²– AN²
PA × PB=T²

Answer 2

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