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!!!

Easy Math Editor

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Comments

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

Ajit Athle - 7 years, 6 months ago

thanx ......

Saksham Aggarwal - 7 years, 6 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$