The chord of the circle x^{2}+y^{2}-4x=0 whic | vedclass

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(D) The equation of the circle is $x^{2}+y^{2}-4x=0$. The center of the circle is $C(2,0)$.Let the chord be bisected at point $M(1,0)$.The line segmen

Vedclass · Accepted Answer

$y=1$

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(D) The equation of the circle is $x^{2}+y^{2}-4x=0$. The center of the circle is $C(2,0)$.Let the chord be bisected at point $M(1,0)$.The line segment connecting the center $C(2,0)$ and the midpoint $M(1,0)$ is perpendicular to the chord.The slope of the line $CM$ is $m_{CM} = \frac{0-0}{1-2} = 0$.Since the chord is perpendicular to $CM$,and $CM$ is a horizontal line (the x-axis),the chord must be a vertical line.$A$ vertical line passing through $(1,0)$ has the equation $x=1$.The slope of this chord is undefined (vertical).We check the options for a line perpendicular to the chord (a vertical line). $A$ line perpendicular to a vertical line must be a horizontal line.Among the options,$y=1$ is a horizontal line.Therefore,the chord is perpendicular to the line $y=1$.

The chord of the circle $x^{2}+y^{2}-4x=0$ which is bisected at $(1,0)$ is perpendicular to the line

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