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Circles Creating Trouble!

Two distinct chords drawn from the point (p,q) on the circle x^{2} + y^{2} = px+qy ; where pq is not 0.. are bisected by the x-axis. Then..... (a) p^{2}= 8q^{2} (b) p^{2}> 8q^{2} (c)p^{2}> 8q^{2} (d) p(mod)= q(mod)

i've tried this question for almost an hour but couldn't get to the answer so please do help!!!

Note by Saksham Aggarwal
3 years, 11 months ago

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Let the chord originating at (p,q) intersect the x-axis in (a,0). It's other extreme is then (2a-p,-q) which must lie on the given circle viz. x²+y²=px+qy or (a-p)²+(-q)² =p(2a-p)+q(-q). Simplify and solve for 'a' giving a=(3p+(p²-8q²)^(1/2))/4 or (3p -(p²-8q²)^(1/2))/4. For real 'a' with two distinct values, we must have p²>8q²

One Top - 3 years, 11 months ago

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thanx ......

Saksham Aggarwal - 3 years, 11 months ago

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