If two chords,each bisected by the x-axis,can | vedclass

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(C) The equation of the circle is $x^2 + y^2 - ax - \frac{b}{2}y = 0$.Let the chord pass through $P(a, b/2)$ and be bisected by the $x$-axis at some p

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$a^2 > 2b^2$

Vedclass · Answer

(C) The equation of the circle is $x^2 + y^2 - ax - \frac{b}{2}y = 0$.Let the chord pass through $P(a, b/2)$ and be bisected by the $x$-axis at some point $Q(h, 0)$.The midpoint of the chord is $Q(h, 0)$. The line segment $PQ$ is a chord of the circle.The center of the circle is $C(a/2, b/4)$.The line $CQ$ must be perpendicular to the chord $PQ$.The slope of $PQ$ is $m_1 = \frac{b/2 - 0}{a - h} = \frac{b}{2(a - h)}$.The slope of $CQ$ is $m_2 = \frac{0 - b/4}{h - a/2} = \frac{-b/4}{h - a/2} = \frac{-b}{4h - 2a}$.Since $PQ \perp CQ$,$m_1 	imes m_2 = -1$.$\frac{b}{2(a - h)} 	imes \frac{-b}{4h - 2a} = -1$$\frac{-b^2}{2(a - h) 	imes 2(2h - a)} = -1$$b^2 = 4(a - h)(2h - a) = 4(2ah - a^2 - 2h^2 + ah) = 4(-2h^2 + 3ah - a^2) = -8h^2 + 12ah - 4a^2$.$8h^2 - 12ah + 4a^2 + b^2 = 0$.For two distinct chords to exist,there must be two distinct values of $h$.Thus,the discriminant $D > 0$.$D = (-12a)^2 - 4(8)(4a^2 + b^2) > 0$$144a^2 - 32(4a^2 + b^2) > 0$$144a^2 - 128a^2 - 32b^2 > 0$$16a^2 > 32b^2$$a^2 > 2b^2$.

If two chords,each bisected by the $x$-axis,can be drawn to the circle $2(x^2 + y^2) - 2ax - by = 0$ $(a \ne 0, b \ne 0)$ from the point $P(a, b/2)$,then:

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