# Intersection between a sphere and a plane and parametrization

I am given that a circle is formed when the unit sphere $x^2+y^2+z^2=1$ intersects the plane $x+y+z=0$. I would like to find the equation of that circle using cylindrical coordinates so that I later can parametrize the equation of the circle. Here is the illustration: $$\left\{ \begin{array}{l} {x^2} + {y^2} + {z^2} = 1\\ x + y + z = 0 \end{array} \right.$$

1. Converting to cylindrical coordinates

Since I am only familiar with cylindrical coordinates (not the way equations should be expressed), I thought that the natural thing would be to simply substitute $x=r\cos(\theta)$ and $y=r\sin(\theta)$, $z=z$ in both equations. Then, I got the following: $$\left\{ \begin{array}{l} {r^2} + {z^2} = 1\\ r(\cos \theta + \sin \theta )+z = 0 \end{array} \right.$$

2. Finding the parametrization

As I know, there are infinitly many parametrizations, so there is no "right one". So, since I have a representation of the system in cylindrical coordinates, I simply substitute $z=-r(\cos \theta + \sin \theta )$ into the first eq. so that I get:

$$\begin{array}{l} {r^2} + {( - r(\cos \theta + \sin \theta ))^2} = 1\\ {r^2} + {r^2}({\cos ^2}\theta + {\sin ^2}\theta + 2\cos \theta \sin \theta ) = 1\\ {r^2}(1 + 1 + \sin 2\theta ) = 1\\ r = \sqrt {\frac{1}{{1 + 1 + \sin 2\theta }}} \end{array}$$

In a similar way, I would find $z$. Now the thing that worries me is the square root sign. I now that sometimes, having a square root sign involved will cause paramatrization of only parts of the curve.

My question is, am I correct in my reasoning and do I have the parametrization for the entire surface?

• I did not check the calculations, but in cylindrical coordinates the radius $r$ is always positive. Since the ambiguity in the solution of the algebraic equation is one between positive and negative values, at least that should not be a problem here. If you get two positive solutions you may be in trouble, but you don't. Feb 3, 2015 at 18:28