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

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

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

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

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

answer is D.

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Detailed Solution

Let the centre of the circle be $(h, k)$ . Since the circle touches the axis of $y$ . Therefore, radius. $= h .$

The radius of the circle $x^{2} + y^{2} - 6 x - 6 y + 14 = 0$ is 2 and it

has its centre at (3, 3). It is given that the two circles touch each other externally

$∴$ Distance between the centres = Sum of the radii

$\begin{array}{l} \Rightarrow \sqrt{(h - 3)^{2} + (k - 3)^{2}} = h + 2 \\ \Rightarrow k^{2} - 10 h - 6 k + 14 = 0 \end{array}$

Hence, the locus of ( $h, k$ is $y^{2} - 10 x - 6 y + 14 = 0$ .

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