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  • If two distinct chords drawn from the point(p, q) on the circle <img alt="\mathrm{ x^2+y^2=p x+q y( where \ p q \neq 0)}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B%20x%

If two distinct chords drawn from the point(p, q) on the circle \mathrm{ x^2+y^2=p x+q y( where \ p q \neq 0)} are bisected by the x-axis then
 

Option: 1

\mathrm{p}^2=\mathrm{q}^2


Option: 2

\mathrm{p}^2=8 \mathrm{q}^2


Option: 3

\mathrm{p}^2<8 \mathrm{q}^2


Option: 4

\mathrm{p}^2>8 \mathrm{q}^2


Answers (1)

best_answer

From equation of circle it is clear the circle passes through origin. Let A B is chord of the circle. \mathrm{A \equiv(p, q)} ; c is mid point and coordinate of c is (h, 0) then coordinaties of B are (-\mathrm{p}, 2 \mathrm{~h},-\mathrm{q}) and \mathrm{B} lies on the circle \mathrm{x}^2+\mathrm{y}^2=\mathrm{px}+\mathrm{qy} we have

\begin{aligned} & (-\mathrm{p}+2 \mathrm{~h})^2+(-\mathrm{q})^2=\mathrm{p}(-\mathrm{p}+2 \mathrm{~h})+\mathrm{q}(-\mathrm{q}) \\ & \Rightarrow 2 \mathrm{p}^2+2 \mathrm{~g}^2-6 \mathrm{ph}+4 \mathrm{~h}^2=0 \Rightarrow 2 \mathrm{~h}^2-3 \mathrm{ph}+\mathrm{q}^2+\mathrm{p}^2-0 \end{aligned}

There are given two distinct chords which are bisected at x-axis then there will be two distinct values of hatisfies equation (1)
So discriminant of this quadrict equation must be >0.

\begin{aligned} & \Rightarrow \mathrm{D}>0 \Rightarrow(-3 \mathrm{p})^2-4.2\left(\mathrm{p}^2+\mathrm{q}^2\right)>0 \\ & 9 \mathrm{p}^2-8 \mathrm{p}^2-8 \mathrm{q}^2>0 \\ & \mathrm{p}^2-8 \mathrm{q}^2>0 \Rightarrow \mathrm{p}^2>8 \mathrm{q}^2 \end{aligned}
 

Posted by

Sanket Gandhi

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