Confusion 2D and 3D

Question

I don't understand why I obtain a cone in 3D if I plot the function of a circle : x^2+y^2 = 0.

Why I do not get a circle in 2D ?

For example, when I plot y = x, I have a line because it's in two dimensions, when I plot y = 1 or x = 1 it is in one dimension. So why for x^2 + y^2 = 0 it's in three dimensions ?

Same, a sphere equation (eg: (x-1)^2 + (y-1)^2 + (z-1)^2 = 0 ) is in 3D as well as the circle one, so that's my confusion

I hope I didnt say to many ineptitudes...

Answer 1

The equation itself does not determine how many dimensions you are working in. The number of dimensions has to be specified independently. For example, $x=1$ could be graphed in 1 dimension, where it's a point; or 2 dimensions, where it's a line; or 3 dimensions, where it's a plane, etc. Same for your other examples.

Also, the equation $x^2+y^2=0$ in 2 dimensions is not a circle. It is a point, because the only (real) numbers $(x,y)$ that satisfy $x^2+y^2=0$ are $x=0, y=0$. To get a circle, the right hand side needs to be a positive number, for example $x^2 + y^2 = 1$. Same with the sphere.

Answer 2

One of the ways in which you may get more insight, is by thinking about how many degrees of freedom you have for a given equation. For example if you consider the equation $x^{2} + y^{2} = 0$. You can see that the only points $(x,y,z)$ in the three dimensional space that satisfy this equation are all the points for which $x = 0$ and $y = 0$. These are all the points $(0,0,z)$ where $z$ is a real number. Thus in three dimensional space, this equation yields a line.

The reason why you do not get a circle in two dimension is because the general equation for a circle in two dimensions is $x^{2} + y^{2} = r^{2}$, where $r$ is the radius of the circle. The equation you have written is a circle with radius $0$ and hence it is just a point in two dimensions.

If you choose a radius different from $r$ this equation will yield a circle in two dimensions and a cylinder in three dimension. Again, the easiest way to tackle such problems is by closely examining which points in the space satisfy the equations.