Is it possible to derive the implicit equation of a cone $x^2+y^2-z^2=0$ from the circle equation $x^2+y^2=1$, which is the intersection between the cone and the hyperplane $\{(x,y,z)\in\Bbb R^3\,|\,z=1\}$?
When I assume that the circle is on the plane parallel to the $xy$ plane with $z=1$ then every point of the cone could be described by a line that goes through the origin to some point of the circle.
In Cartesian coordinates, if $(x,y,1)$ is a point of the unit circle, then $(x,y,1)\cdot t$, where $t$ is a real number, is a point of the cone.
I hope it is clear what I mean. Thank you
Yes, it is possible. All you have to do is "projectivize" your equation. If $[X:Y:Z]$ are homogeneous coordinates in the projective space, then $(X/Z, Y/Z)$ are the coordinates of the trace in the ordinary plane. If you plug in $x=X/Z$, $y=Y/Z$ in your equation you get the equation of the desired cone.