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The chord of the circle $x^{2}+y^{2}-4 x=0$ which is bisected at $(1,0)$ is perpendicular to the line
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Verified Answer
The correct answer is:
$y=1$
Given, equation of circle
$$
x^{2}+y^{2}-4 x=0 \text {, bisect by }(1,0)
$$
Centre of circle $=(2,0)$
Radius of circle $=2$
From figure, $\mathrm{AB}$ is the chord of circle which bisected by the point $(1,0)$ and perpendicular to the line $\mathrm{y}=1$.
Because it is parallel to x-axis while chord is parallel to $\mathrm{y}$-axis.
$$
x^{2}+y^{2}-4 x=0 \text {, bisect by }(1,0)
$$
Centre of circle $=(2,0)$
Radius of circle $=2$
From figure, $\mathrm{AB}$ is the chord of circle which bisected by the point $(1,0)$ and perpendicular to the line $\mathrm{y}=1$.
Because it is parallel to x-axis while chord is parallel to $\mathrm{y}$-axis.
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