From the origin,chords are drawn to the circl | vedclass

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(A) Let $(x', y')$ be the midpoint of a chord.The equation of the chord with midpoint $(x', y')$ for the circle $x^2 + y^2 - 2x = 0$ is given by $T =

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(A) Let $(x', y')$ be the midpoint of a chord.The equation of the chord with midpoint $(x', y')$ for the circle $x^2 + y^2 - 2x = 0$ is given by $T = S_1$.$x x' + y y' - (x + x') = x'^2 + y'^2 - 2x'$Since the chord passes through the origin $(0, 0)$,we substitute $x = 0$ and $y = 0$ into the equation:$0 + 0 - (0 + x') = x'^2 + y'^2 - 2x'$$-x' = x'^2 + y'^2 - 2x'$$x'^2 + y'^2 - x' = 0$Replacing $(x', y')$ with $(x, y)$,the locus is $x^2 + y^2 - x = 0$,which represents a circle.

From the origin,chords are drawn to the circle $(x - 1)^2 + y^2 = 1$. The locus of the midpoints of these chords is a

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