The curve produced by intersecting a sphere with a phane

Question

Given are the sphere $x^2 + y^2 + z^2 = 1$ and the plane $x + y = 0$.

How can I find a parametrization for the intersection of the two? I found it an ellipse? The plane passes through from the center of sphere. What's wrong?

Answer 1

It's just $y=-x$ and $z^2=1-2x^2$.

Answer 2

Solving $x + y = 0$ for $y$ in terms of $x$ gives $y = -x$, and substituting in the equation of the sphere gives the equation $$2 x^2 + z^2 = 1,$$ which we can regard as defining a curve in the $xz$-plane. By construction, this curve is the projection of the original curve (the intersection of the sphere and the plane) onto that plane, and the projected curve is an ellipse.

The form of this equation suggests writing this curve as $$(\sqrt{2} x)^2 + z^2 = 1,$$ which we may parameterize by $(x(t), y(t))$, where $\sqrt{2} x(t) = \cos t$, $z(t) = \sin t$.

We can now produce a parameterization of the original curve using the fact that for any point on our original curve, $y = -x$: $$x(t) = \frac{1}{\sqrt{2}} \cos t, \qquad y(t) = - \frac{1}{\sqrt{2}} \cos t, \qquad z(t) = \sin t .$$ Like any nondegenerate intersection of a plane with a sphere, our curve is a circle.