I have three points: A(85, 85, 0), B(-85, -85, 0) and C(0, 0, 30). I must find the equation of the arch that starts from A, finishes in B and goes through C.
Could you help me? I found something called catenary but it's something that overtakes my basic knowledge. Thank you.
Your two examples are surfaces, not curves. In fact (roughly speaking) any implicit equation of the form $f(x,y,z)=0$ will yield a surface.
One arch-like surface that will work is a parabolic one related to the previous answer:
$$ z = 30 \left(1-\frac{y^2}{7225}\right)$$
This is an arch-shaped surface that runs parallel to the x-axis. Similarly
$$ z = 30 \left(1-\frac{x^2}{7225}\right)$$ is an arch-shaped surface that runs parallel to the x-axis. Finally, the arch surface
$$z = 30 \left(1-\frac{(x+y)^2}{28900}\right)$$
is somehow "nicer" because it's symmetric in $x$ and $y$.
Taking the intersection of any of these three with the plane $x=y$ will give a curve that meets your specifications.