Question

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

Answer 1

★ GEOMETRY CIRCLES ★

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Given:-

PAB is the secant and PT is the tangent to the circle with centre O.

Construction:-

Draw ON ⊥ AB

Proof:-

Here, AN = BN
Also, we have, PT² = PA × PB [By tangent secant property] ---- (1)

Consider ΔONA,
So, 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

Thus, PT² = PN² – AN² ----(3)

From (1) and (3), We get,

∴ PA × PB = PN² – AN²

--Please refer to the attached image for clarification of diagram and points--

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