consider 3D space,
- $x^2+y^2+z^2=1$ and $y^2+z^2=1$
- $x^2+y^2+z^2=1$ and $x=0$
- $y^2+z^2=1$ and $x=0$
i think they are all the same thing , indicating 2D circle $y^2+z^2=1$ but since there are no solutions, i have to check it out
- will be a sphere and a cylinder so that there are intersections which is only a circle
- will be a sphere and a plane so that there are a circle
- will be a cylinder and a plane so thay there are a circle , i think
Inserting the second equation of 1. into the first, derives $x^2=0$ and thus $x=0$. Thence 1. implies both 2. and 3.
Inserting the second equation from 2. into the first, derives both 1. and 3.
Inserting the second equation from 3. into the first (simply adding 0 to either side (or rather its square)), derives both 1. and 2.
Thus you'll prove their equivalences.
--- rk