The locus of the centre of a circle which tou | vedclass

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(D) Let the center of the circle be $(h, k)$. Since it touches the $y$-axis,its radius $r = |h|$.The given circle is $x^2 + y^2 - 6x - 6y + 14 = 0$. I

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

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(D) Let the center of the circle be $(h, k)$. Since it touches the $y$-axis,its radius $r = |h|$.The given circle is $x^2 + y^2 - 6x - 6y + 14 = 0$. Its center $C_1 = (3, 3)$ and radius $r_1 = \sqrt{3^2 + 3^2 - 14} = \sqrt{18 - 14} = 2$.Since the circles touch externally,the distance between centers $C_1(3, 3)$ and $C_2(h, k)$ is equal to the sum of their radii:$C_1C_2 = r_1 + r_2$$\sqrt{(h-3)^2 + (k-3)^2} = 2 + |h|$Squaring both sides:$(h-3)^2 + (k-3)^2 = (h+2)^2$$h^2 - 6h + 9 + k^2 - 6k + 9 = h^2 + 4h + 4$$k^2 - 6k - 10h + 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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