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Using coordinate geometry prove that for a circle and external point P PA.PB=PT^2?
where P is any external point s.t. PAB is a secant & PT is a tangent
Its a well known result of geometry
1 Answer
- MadhukarLv 71 decade agoFavorite Answer
Let AB be the chord of a circle of radius R and P be any external point of the circle on AB.
Choose the coordinate axes such that origin is at the centre of the circle and x-axis is parallel to the chord AB.
Let A = (- Rcosθ, Rsinθ), B = (Rcosθ, Rsinθ) and P = (a, Rsinθ)
The equation of the circle is x^2 + y^2 = R^2
PT^2 = a^2 + R^sin^2 θ - R^2 = a^2 - R^2cos^2 θ ... (1)
PA = [(a + Rcosθ)^2]^(1/2) = a + Rcosθ and
PB = [(a - Rcosθ)^2]^(1/2) = a - Rcosθ
=> PA * PB = (a + Rcosθ) * (a - Rcosθ) = a^2 - R^2cos^2 θ ... (2)
From (1) and (2),
PA*PB = PT^2.
