How to parametrise a circle in a plane I've been studying for an exam and in one of the exams it asks me the parametrise the unit circle in $R^3$ which lines in the plane $$x+y=0$$.

I know how to parametrise a normal circle, but I don't really have any idea how to parametrise it in a plane like that. Any help would be appreciated, thanks
The unit sphere has equation $x^2+y^2+z^2=1$, its intersection with the plane is the circle you are looking for. So you have $$y=-x$$ and $$2x^2+z^2=1$$. You can now use polar coordinates so put $x=-y={1 \over \sqrt 2} \cos t$ and $z=\sin t$ so that  $$2{1 \over  2} \cos^2 t+ \sin^2 t=1$$ is identically verified.
Here is what I think. The way to derive a circle is to set up the identity of distance, i.e. the square of the distance of a point on the circle to its center (1 for unit circle) is equal to the sum of squares(Pythagorean theorem). If you are given the points on $x+y=0$ plane, then the setting of such points would be limited to $(x,-x,z)$. Do the same thing. Since now you are guaranteed the points in a plane, it is for sure a circle, not a sphere.