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  • If two distinct chords, drawn from the point on the circle <img alt="\mathrm{x^2+y^

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

Option: 1

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


Option: 2

\mathrm{ p=8 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 that circle passes through origin. Let A B is chord of the circle.

\mathrm{A \equiv(p, q) \cdot C}is mid-point and co-ordinate of C is \mathrm{(h, 0)}
Then coordinates of B are \mathrm{(-p+2 h,-q)} and 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 \mathrm{p}^2+4 \mathrm{~h}^2-4 \mathrm{ph}+\mathrm{q}^2=-\mathrm{p}^2+2 \mathrm{ph}-\mathrm{q}^2 \\ & \Rightarrow 2 \mathrm{p}^2+2 \mathrm{q}^2-6 \mathrm{ph}+4 \mathrm{~h}^2=0 \\ & \Rightarrow 2 \mathrm{~h}^2-3 \mathrm{ph}+\mathrm{p}^2+\mathrm{q}^2=0 \quad \quad \quad \quad \quad \quad(1) \end{aligned}

there are given two distinct chords which are bisected at x-axis then, there will be two distinct values of h satisfying (1).

So discriminant of this quadratic equation must be >0.

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

\mathrm{\Rightarrow p^2>8 q^2}, therefore (D) is the answer.

\frac{\mathrm{t}-2}{2}=\mathrm{h}, \frac{\mathrm{t}+2}{2}=\mathrm{k} \Rightarrow \mathrm{k}=\mathrm{h}+2

Posted by

HARSH KANKARIA

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