The locus of the centre of a circle which tou | 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 $r = |h|$. Assuming $h > 0$,we hav

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$y^2 - 10x - 6y + 14 = 0$

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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 $r = |h|$. Assuming $h > 0$,we have $r = h$.The given circle is $x^2 + y^2 - 6x - 6y + 14 = 0$. Its centre $C_1$ is $(3, 3)$ and its radius $R_1 = \sqrt{3^2 + 3^2 - 14} = \sqrt{18 - 14} = \sqrt{4} = 2$.Since the circles touch externally,the distance between their centres $C(h, k)$ and $C_1(3, 3)$ is equal to the sum of their radii:$CC_1 = R_1 + r$$\sqrt{(h - 3)^2 + (k - 3)^2} = 2 + h$Squaring both sides:$(h - 3)^2 + (k - 3)^2 = (2 + h)^2$$h^2 - 6h + 9 + k^2 - 6k + 9 = 4 + 4h + h^2$$k^2 - 6k + 18 - 6h = 4 + 4h$$k^2 - 10h - 6k + 14 = 0$Replacing $(h, k)$ with $(x, y)$,the locus is $y^2 - 10x - 6y + 14 = 0$.

The locus of the centre of a circle which touches externally the circle $x^2 + y^2 - 6x - 6y + 14 = 0$ and also touches the $y$-axis,is given by the equation

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