# Procedure of solving a two variables polynomial?

in case if it needs to be said for more information, this is for a level curves (contour lines) equation.

Well, the problem is that from what I know if I want to graph, I would need one dependent variable and the independent variables or the constant.

But in this equation:

$$z - 1 = (x-2)^2 + (y-2)^2$$

Z is a real number constant and not a variable (It can be called as K) in level curves. Z or K can be 1, 2, 3, etc.

So as you can see, I would have two "x" and two "y" if I solve the polynomial and this is the problem, I can't graph with two same variables (From what I know, correct me if I'm wrong).

$$z - 1 = x^2 - 4x + 4 + y^2 - 4y + 4$$

I'm trying to achieve is to make x or y as the dependent variable to graph in 2D plane.

Thanks for the answer. Hope you can understand.

The graph of $$(x-2)^2 + (y-2)^2 = z-1$$ is the circle with centre $$(2,2)$$ and radius $$\sqrt{z-1}$$ if $$z > 1$$, the single point $$(2,2)$$ if $$z=1$$, and empty if $$z < 1$$. If you insist on plotting with one variable as dependent and the other independent, then in the $$z>1$$ case you can write $$y = 2 \pm \sqrt{z - 1 - (x-2)^2},\ 2 - \sqrt{z-1} \le x \le 2 + \sqrt{z-1}$$ With $$+$$ you get the top half of the circle, with $$-$$ you get the bottom half.