Question

If two distinct chords drawn from the point (p, q) on the circle $x^2+y^2-px-qy=0(wherepq\ne 0)$ are bisected by the x-axis, then

• A. $p^2=q^2$
• B. $p^2=8q^2$
• C. $p^2<8q^2$
• D. $p^2>8q^2$

Answer 1

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

Let B(h, 0) be the midpoint of the chord drawn from point A(p, q).
Also, the center of the circle is $C(\frac{p}{2},\frac{q}{2})$.
Then, we have $BC\perp AB$. Therefore,
$\frac{\left(\frac{q}{2}\right)-0}{\left(\frac{p}{2}\right)-h}\left(\frac{q-0}{p-h}\right)=-1...\text{(Product of slopes of two perpendicular lines},m_1.m_2=-1)\therefore \left(\frac{q}{p-2h}\right)\left(\frac{q-0}{p-h}\right)=-1\therefore 2h^2-3ph+p^2+q^2=0$
Since two such chords exist, the above equation must have two distinct real roots, i.e.,
Discriminant > 0
$\therefore 9p^2-8(p^2+q^2)>0or{p}^2>8q^2$