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

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(C) The given circle is $(x - 1)^2 + y^2 = 1$,which expands to $x^2 + y^2 - 2x = 0$.Let $(h, k)$ be the midpoint of a chord.The equation of a chord wi

Vedclass · Accepted Answer

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

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(C) The given circle is $(x - 1)^2 + y^2 = 1$,which expands to $x^2 + y^2 - 2x = 0$.Let $(h, k)$ be the midpoint of a chord.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,the equation of the chord is $xh + yk - (x + h) = h^2 + k^2 - 2h$.Since the 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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