Locus of the centre of the circle which touches circle $x^2+y^2-6x-6y+14=0$ externally and also y-axis is

The given circle is $x^2+y^2-6x-6y+14=0$ 

C$C_1\left(3,3\right),r_1=\sqrt{9+9-14}=2$

Suppose that $\left(x_1,y_1\right)$ be any point on the locus, so it is centre of the other circle.

Since it touches the $y-$ axis radius is $\left|x_1\right|$

And both circles touch each other externally, so that $c_1{c}_2=r_1+r_2$

Hence

${\left(x_1-3\right)}^2+{\left(y_1-3\right)}^2={\left(\left|x_1\right|+2\right)}^2$

Simplify and then generalize the above equation

Therefore, the required locus is $y^2-6y-10x+14=0$