# 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?

• Hint: The radius to the point of tangency is perpendicular to the tangent line. – quasi Aug 8 at 3:35

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.

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 $$y_0=3$$.

• I didn't think of it! But sadly, the general equation it's not given, I have to find it. Maybe I need to improve my question. – Luis Victoria Aug 8 at 3:48