Answer
$(x+3)^2+(y-1)^2=3^2$.
Work Step by Step
The general equation for a circle with radius $r$ and centre $(h,k)$ is: $(x-h)^2+(y-k)^2=r^2$.
The distance formula from $P_1(x_1,y_1)$ to $P_2(x_2,y_2)$ is $d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$.
If the circle is tangent to the y-axis, its radius is the absolute value of is x-coordinate, hence here: $r=3.$
Also $C=(-3,1)$, hence our equation: $(x+3)^2+(y-1)^2=3^2$.
