The locus of the centre of the circles passin | vedclass

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(C) Let the centre of the circle be $(h, k)$.Since the circle passes through the origin $(0, 0)$,its radius $R$ is given by $R^2 = h^2 + k^2$.The equa

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a parabola

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(C) Let the centre of the circle be $(h, k)$.Since the circle passes through the origin $(0, 0)$,its radius $R$ is given by $R^2 = h^2 + k^2$.The equation of the circle is $(x-h)^2 + (y-k)^2 = h^2 + k^2$,which simplifies to $x^2 + y^2 - 2hx - 2ky = 0$.The circle cuts a chord of length $2$ units on the line $x=1$. The length of the chord is given by $2\sqrt{R^2 - d^2} = 2$,where $d$ is the perpendicular distance from the centre $(h, k)$ to the line $x=1$.Thus,$\sqrt{R^2 - d^2} = 1$,or $R^2 - d^2 = 1$.Here,$R^2 = h^2 + k^2$ and $d = |h-1|$.Substituting these values,we get $(h^2 + k^2) - (h-1)^2 = 1$.$h^2 + k^2 - (h^2 - 2h + 1) = 1$$h^2 + k^2 - h^2 + 2h - 1 = 1$$k^2 + 2h - 1 = 1$$k^2 = 2 - 2h$Replacing $(h, k)$ with $(x, y)$,the locus is $y^2 = 2(1-x)$,which represents a parabola.

The locus of the centre of the circles passing through the origin and cutting off a chord of length $2$ units on the line $x=1$ is

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