The locus of the centre of the circle touchin | vedclass

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(C) Let the centre of the circle be $(h, k)$.Since the circle touches the $x$-axis,the radius $r$ of the circle is equal to $|k|$.Since the circle pas

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a parabola with focus at $(-1, 1)$

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(C) Let the centre of the circle be $(h, k)$.Since the circle touches the $x$-axis,the radius $r$ of the circle is equal to $|k|$.Since the circle passes through the point $(-1, 1)$,the distance from the centre $(h, k)$ to the point $(-1, 1)$ must be equal to the radius $r$.Thus,$\sqrt{(h - (-1))^2 + (k - 1)^2} = |k|$.Squaring both sides,we get $(h + 1)^2 + (k - 1)^2 = k^2$.$(h + 1)^2 + k^2 - 2k + 1 = k^2$.$(h + 1)^2 = 2k - 1$.Replacing $(h, k)$ with $(x, y)$,we get $(x + 1)^2 = 2y - 1$,which can be rewritten as $(x + 1)^2 = 2(y - 1/2)$.This is the equation of a parabola of the form $(x - h)^2 = 4a(y - k)$,where the focus is at $(h, k + a)$.Here,$4a = 2$,so $a = 1/2$. The vertex is $(-1, 1/2)$.The focus is $(-1, 1/2 + 1/2) = (-1, 1)$.Therefore,the locus is a parabola with focus at $(-1, 1)$.

The locus of the centre of the circle touching the $x$-axis and passing through the point $(-1, 1)$ is

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