Equation of a circle tangent to the $y$ axis The problem is:

Find the general equation of a circumference with center at $(-4,3)$ and tanget to the $y$ axis

I know that calculating the distance between the center and any point on the circumference gives me the radios. And then the general equation is pretty straight forward.
But I'm stuck, I tried to find exactly the point were the circumference meets the $y$ axis. So the point has to be of the form $(0,y_0)$ . I tried to calculate the distance directly of this point and the center of the circumference but nothing came out of it.
The final result should be:
$$x^2+y^2+8x-6y+9=0$$
I found a lot of mistakes in the book I'm using and I'm suspecting that this problem is missing something. Does anyone have any suggestions?
 A: The distance from $(-4,3)$ to the $y$-axis is $4$. This is because the $y$-axis is the line where $x = 0$, so the tangent line is a vertical line and the horizontal distance to it (as quasi's question comment indicates, the radius line to the point of tangency & the tangent line are perpendicular to each other) from the center is $|-4 - 0| = 4$, occurring at the point $(0,3)$. Thus, the radius is $4$. The general equation of a circle at center $(x_0,y_0)$ with radius $r$ (e.g., as given in the Equations section of Wikipedia's Circle article) is
$$(x - x_0)^2 + (y - y_0)^2 = r^2 \tag{1}\label{eq1}$$
Using the known values gives
$$(x+4)^2 + (y-3)^2 = 4^2 \tag{2}\label{eq2}$$
Expanding & simplifying gives
$$x^2 + 8x + 16 + y^2 - 6y + 9 = 16 \implies x^2 + y^2 + 8x - 6y + 9 = 0 \tag{3}\label{eq3}$$
As you can see, this matches the problem solution you wrote about in your question.
A: Since the tangent point is $(0,y_0)$, you may just plug it into the equation
$$x^2+y^2+8x-6y+9=0$$
to get 
$$y_0^2-6y_0+9=0$$
Then, $y_0=3$.
Suppose, the equation of the circle is unavailable yet, you could still argue that the tangent point $(0,y_0)$ is the intersection between the $y$-axis and a horizontal line going through the center of the circle. Since the center is (-4,3), the horizontal line must be $y=3$. Thus $y_0=3$.
