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Q.
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
Conic Sections
Solution:
Let the centre of the circle be $\left(h, k\right)$. Since the circle touches the axis of $y$.
$\therefore $ Its radius will be $r_{1} = h$.
Centre of the other given circle is $\left(3,3\right)$ and radius, $r_{2} = 2$.
Since circles touch externally so distance between centres
$=r_{1}+r_{2}=h+2$
$\Rightarrow \, \left(h-3\right)^{2}+\left(k-3\right)^{2}=\left(h+2\right)^{2}$
$\Rightarrow \, k^{2}-10h-6k+14=0$
$\therefore $ Required locus is $y^{2} - 10x -6 y+ 14 = 0$.