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

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(C) The equation of the circle is $(x - 1)^2 + y^2 = 1$,which simplifies to $x^2 - 2x + 1 + y^2 = 1$,or $x^2 + y^2 - 2x = 0$.Let $(h, k)$ be the midpo

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$x^2 + y^2 - x = 0$

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(C) The equation of the circle is $(x - 1)^2 + y^2 = 1$,which simplifies to $x^2 - 2x + 1 + y^2 = 1$,or $x^2 + y^2 - 2x = 0$.Let $(h, k)$ be the midpoint of a chord passing through the origin $(0, 0)$.The equation of a chord with midpoint $(h, k)$ is given by $T = S_1$,where $T = xh + yk - (x + h)$ and $S_1 = h^2 + k^2 - 2h$.So,$xh + yk - x - h = h^2 + k^2 - 2h$.Since this chord passes through the origin $(0, 0)$,we substitute $x = 0$ and $y = 0$ into the equation:$0(h) + 0(k) - 0 - h = h^2 + k^2 - 2h$.$-h = h^2 + k^2 - 2h$.$h^2 + k^2 - h = 0$.Replacing $(h, k)$ with $(x, y)$,the locus is $x^2 + y^2 - x = 0$.

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

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