Question

from an external point P, a tangent PT and a secant PAB is drawn to a circle with centre O. ON is perpendicular on the chord AB. Prove (i) PA.PB = PN² – AN² (ii) PN² – AN² = OP² – PT² (iii) PA.PB = PT²

Answer 1

Given PAB is the secant and PT is the tangent to the circle with centre O
Draw ON ⊥�AB
Therefore, AN = BN
Also we have PT2�= PA × PB [By tangent secant property] ---- (1)
Consider ΔONA
OA2�= ON2�+ AN2� ---- (2)
Similarly in�ΔPTO
OP2�= 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

Answer 2

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