Question

If two distinct chords, drawn from the point (p, q) on the circle $x^{2} + y^{2} = px + qy$ (where p, q ≠ 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

Answer: (D)

$p^{2} > 8q^{2}$

Answer 2

Let (h, 0) be a point on x-axis, then the equation of chord whose mid-point is (h, 0) will be

$xh - \frac{1}{2}p(x + h) - \frac{1}{2}q(y + 0) = h^{2} - ph$.

This passes through (p, q),

hence $p h - \frac { 1 } { 2 } p ( p + h ) - \frac { 1 } { 2 } q \cdot q = h ^ { 2 } - p h$

$\Rightarrow ph - \frac{1}{2}p^{2} - \frac{1}{2}ph - \frac{1}{2}q^{2} = h^{2} - ph \Rightarrow h^{2} - \frac{3}{2}ph + \frac{1}{2}(p^{2} + q^{2}) = 0$;

$\because$ h is real, hence $B^{2} - 4AC > 0$

$\therefore$ $\frac{9}{4}p^{2} - 4.\frac{1}{2}(p^{2} + q^{2}) > 0 \Rightarrow 9p^{2} - 8(p^{2} + q^{2}) > 0 \Rightarrow p^{2} - 8q^{2} > 0 \Rightarrow p^{2} > 8q^{2}$