# Drawing level curves on x-y axes

Suppose I sketched the domain of a function $$f(x, y) = \sqrt{(x-1)^2 + (y-2)^2 - 1}$$ on the x-y plane. There's no problem with that. But suppose on that same set of x-y axes I wanted to add (by hand) some level curves of the function for $$z = \sqrt{3}, \sqrt{7}$$ and $$\sqrt{12}$$ (Feel free to use any $$\sqrt{}$$ values).

If we allow $$z = k = (x-1)^2 + (y-2)^2 ≥ 1$$ I could maybe manoeuvre $$x = y-1$$ which yields $$2(y-2)^2 ≥ 1$$

If this is correct, what steps are required next to input and graph the z-values?

I'm not sure if I'm answering your question, but drawing level curves by hand for this function is convenient by considering the square of the function $$z^2=(x-1)^2 + (y-2)^2 - 1$$ So, we have a constant $$c=r^2=z^2+1$$ where $$r$$ is the radius of a circular level curve in the $$xy$$-plane with center $$(x,y)=(1,2)$$. Hence, for $$z=\sqrt{3}$$ we have $$c=4$$ corresponding to a radius $$r=2$$.

Why not use ubiquitous plotting programs rather than "by hand"?

Such a 3D plot shows the structure extremely well.

But if you're required to lose the information (such as the relative heights of contours), you can easily simplify such graphs:

All the above took no more than one minute in software. I wonder how long the OP will work on doing such plots "by hand" (and not making a mistake) and whether all that hand work really helps.

• To demonstrate understanding in regards to the function I suppose Jan 5, 2021 at 2:53
• I, frankly, don't believe that is the way to understanding. See this: ted.com/talks/… Jan 5, 2021 at 3:08
• Nevertheless, it's required to be within the x-y plane only Jan 5, 2021 at 5:08