Question

Two distinct chords drawn from the point $(p,q)$ on the circle $x^2+y^2=px+qy$ are bisected at the $x$-axis. Then,

• A. $\left|p\right|=\left|q\right|$
• B. $p^2=8q^2$
• C. $p^2<8q^2$
• D. $p^2>8q^2$

Answer 1

The correct option is D $p^2>8q^2$

Given circle is $x^2+y^2=px+qy.$

Since the centre of the circle is $(\frac{p}{2},\frac{q}{2}),$

So $(p.q)$ and $(0,0)$ are the end points of a diameter.

As the two chords are bisected by $x$-axis,

the chords will cut the points $(x_1,-q)$ and $(x_2,-q),$ where $x_1,x_2$ are real.

The equation of the line joining these points is $y=-q.$

Solving $y=-q$ and $x^2+y^2=px+qv,$

we get $x^2-px+2q^2=0$

The roots of this equation are $x_1$ and $x_2$.

Since the roots are real and distinct,

$\therefore$ Discriminent $>0$

$\Rightarrow p^2-8q^2>0\Rightarrow p^2>8q^2$.