Procedure of solving a two variables polynomial?

Question

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.

Answer 1

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.

Answer 2

With level curves, you are trying to plot a 3D graph in the 2D plane. One way to do this is to fix z first. Since now we have a single equation of 2 free variables, there are infinitely many solutions to this. The question now is a 2D question. What happens to y when we vary x? The solution curve here can be plotted in the usual way by choosing various values for x and finding the corresponding y. This curve corresponds to a single level curve for the value of z that you first chose.

This procedure can be repeated for multiple values of z to obtain multiple level curves. You are plotting curves in the x-y plane that correspond to fixed values of z.