The locus of the centres of the circles,which | vedclass

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(D) Let the centre of the circle be $(h, k)$ and its radius be $r$. Since the circle touches the $y$-axis,the radius $r = |h| = h$ (as it lies in the

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$y = \sqrt{1 + 2x}, x \ge 0$

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(D) Let the centre of the circle be $(h, k)$ and its radius be $r$. Since the circle touches the $y$-axis,the radius $r = |h| = h$ (as it lies in the first quadrant,$h > 0$). Since the circle touches $x^2 + y^2 = 1$ externally,the distance between the centres is equal to the sum of the radii: $\sqrt{h^2 + k^2} = r + 1 = h + 1$. Squaring both sides,we get: $h^2 + k^2 = (h + 1)^2$ $h^2 + k^2 = h^2 + 2h + 1$ $k^2 = 2h + 1$. Replacing $(h, k)$ with $(x, y)$,the locus is $y^2 = 2x + 1$,or $y = \sqrt{2x + 1}$ for $x \ge 0$.

The locus of the centres of the circles,which touch the circle $x^2 + y^2 = 1$ externally,also touch the $y$-axis and lie in the first quadrant is

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