If two distinct chords drawn from the point ( | vedclass

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(D) Let the point on the $x$-axis be $(h, 0)$. The equation of the chord with midpoint $(h, 0)$ is given by $T = S_1$.The equation of the circle is $x

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

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(D) Let the point on the $x$-axis be $(h, 0)$. The equation of the chord with midpoint $(h, 0)$ is given by $T = S_1$.The equation of the circle is $x^{2} + y^{2} - px - qy = 0$.The equation of the chord is $xh + y(0) - \frac{p}{2}(x + h) - \frac{q}{2}(y + 0) = h^{2} + 0^{2} - ph - q(0)$.$xh - \frac{p}{2}(x + h) - \frac{qy}{2} = h^{2} - ph$.Since this chord passes through $(p, q)$,we substitute $x = p$ and $y = q$:$ph - \frac{p}{2}(p + h) - \frac{q^{2}}{2} = h^{2} - ph$.Multiplying by $2$: $2ph - p^{2} - ph - q^{2} = 2h^{2} - 2ph$.$2h^{2} - 3ph + p^{2} + q^{2} = 0$.Since there are two distinct chords,the quadratic equation in $h$ must have two distinct real roots,so the discriminant $D > 0$.$D = (-3p)^{2} - 4(2)(p^{2} + q^{2}) > 0$.$9p^{2} - 8p^{2} - 8q^{2} > 0$.$p^{2} - 8q^{2} > 0 \Rightarrow p^{2} > 8q^{2}$.

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

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