If two distinct chords,drawn from the point ( | vedclass

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(D) The equation of the circle is ${x^2} + {y^2} - px - qy = 0$. Let the midpoint of a chord be $(h, 0)$ on the $x$-axis. The equation of the chord wi

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

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(D) The equation of the circle is ${x^2} + {y^2} - px - qy = 0$. Let the midpoint of a chord be $(h, 0)$ on the $x$-axis. The equation of the chord with midpoint $(x_1, y_1)$ is $T = S_1$,where $T = xx_1 + yy_1 - \frac{p}{2}(x + x_1) - \frac{q}{2}(y + y_1)$ and $S_1 = x_1^2 + y_1^2 - px_1 - qy_1$. Substituting $(x_1, y_1) = (h, 0)$,we get: $xh - \frac{p}{2}(x + h) - \frac{q}{2}y = h^2 - ph$. Since the 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 $2$: $2ph - p^2 - ph - q^2 = 2h^2 - 2ph$. Rearranging terms: $2h^2 - 3ph + p^2 + q^2 = 0$. For two distinct chords to exist,the quadratic equation in $h$ must have two distinct real roots. Thus,the discriminant $D > 0$: $D = (-3p)^2 - 4(2)(p^2 + q^2) > 0$. $9p^2 - 8p^2 - 8q^2 > 0$. $p^2 - 8q^2 > 0$. Therefore,$p^2 > 8q^2$.

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

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