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

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(D) Let the center of the required circle be $C_1 = (h, k)$. Since it touches the $y$-axis,its radius $r_1 = |h|$.For the given circle $x^{2} + y^{2}

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

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(D) Let the center of the required circle be $C_1 = (h, k)$. Since it touches the $y$-axis,its radius $r_1 = |h|$.For the given circle $x^{2} + y^{2} - 6x - 6y + 14 = 0$,the center $C_2 = (3, 3)$ and radius $r_2 = \sqrt{3^{2} + 3^{2} - 14} = \sqrt{9 + 9 - 14} = \sqrt{4} = 2$.Since the circles touch externally,the distance between their centers is $C_1 C_2 = r_1 + r_2$.Thus,$\sqrt{(h - 3)^{2} + (k - 3)^{2}} = |h| + 2$.Squaring both sides,$(h - 3)^{2} + (k - 3)^{2} = (|h| + 2)^{2}$.$h^{2} - 6h + 9 + k^{2} - 6k + 9 = h^{2} + 4|h| + 4$.$k^{2} - 6k + 14 = 6h + 4|h|$.Assuming $h > 0$,$k^{2} - 6k + 14 = 10h$.Replacing $(h, k)$ with $(x, y)$,the locus is $y^{2} - 6y + 14 = 10x$,or $y^{2} - 10x - 6y + 14 = 0$.

The locus of the center of a circle which touches the circle $x^{2} + y^{2} - 6x - 6y + 14 = 0$ externally and also touches the $y$-axis is:

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