Can the intersection of an ellipsoid and a plane be a circle? What conditions are necessary for that to happen?

Question

I happened to be working on a problem that tried to mix linear algebra and statistics and the question in the title came to mind.

The goal was to determine the subset of $Y \subset \mathbb R^3$ corresponding to every element $y \in Y$ that have the same expected value and standard deviation of a given element $\overrightarrow x = (x_1,x_2,x_3) \in \mathbb R^3$ for a given $P = (p_1,p_2,p_3)$ such that $0 \leq p_i \leq 1$ and $\sum_{i=1}^3 p_i = 1$.

The expected value and standard deviation were defined respectively as:

$$ \left\{ \begin{array}{} E(\overrightarrow x) = \displaystyle \sum_{i=1}^3 p_ix_i \\ \sigma(\overrightarrow x) = \displaystyle \sqrt{\sum_{i=1}^3 p_i\Bigl(x_i-E(\overrightarrow x)\Bigl)^2} \end{array} \right. $$

I understand that $Y$ must be the intersection of a plane and an ellipsoid, but curiosity made me interested in trying to find the elements $P$ such that $Y$ would happen to be a disk in the plane (from the first condition defining $Y$) that intersects the ellipsoid (from the second condition defining $Y$), and thus came about the question below:

Given an ellipsoid with distinct principal semi-axes, is it generally (or ever) the case that one may find a plane whose intersection with the ellipsoid determines a circle in that plane? If so, under what conditions does that happen?

Thanks in advance. I apologize for any mistakes I might make, this is my first time around.

Answer 1

Place the ellipsoid with its center at the origin and its short axis on the $x$-axis, its long axis on the $y$-axis, and its middle axis on the $z$-axis. The intersection with a plane through the $z$-axis is always an ellipsis with one of its axes on the $z$-axis. For the $xz$-plane, the longer axis is on $z$, for the $yz$ plane, it is the shorter axis. By ontinuity, for some in-between plane, the axes have the same length, i.e., the ellipsis becomes a circle.

Answer 2

I propose here an approach that shows that one can build circular sections for an immense family of ellipsoids.

Let us consider WLOG, in order to keep simple equations, that the circle is the unit circle $(U)$ positionned in the (horizontal) x-y plane.

Consider the unit sphere $(S)$ with equation

$$\underbrace{x^2+y^2+z^2-1}_{f(x,y,z)}=0 \tag{1}$$

Let us define an arbitrary non horizontal direction by a vector $V(a,b,1)$.

Let $(C)$ be the slant cylinder whose intersection with the horizontal plane is $(U)$, and whose generating lines have the direction defined by $V$, i.e., with parametric representation $(x(u,v),y(u,v),z(u,v))$ given by:

$$\begin{cases}x&=&\cos(u)+av\\y&=&\sin(u)+bv\\z&=&v\end{cases}, \tag{2}$$

yielding the implicit equation:

$$\underbrace{(x-az)^2+(y-bz)^2-1}_{g(x,y,z)}=0 \tag{2'}$$

Then, the following equation, combining in a barycentrical manner the equations of the sphere and the hyperboloid:

$$h(x,y,z) \ := \ (1-t) f(x,y,z) \ + \ t g(x,y,z) \ \ \ \ \ 0 < t < 1 \tag{3}$$

is that of an ellipsoid with an increasingly longer semi major axis, in fact tending to $\infty$, when $t$ tends to $1$. of course, this ellipsoid crosses the x-y plane along unit circle $(U)$.

This is what is called a (part of a) pencil of quadrics.

Remark: the fact that (3) represents an ellipsoid can be checked by establishing the positive definiteness of the associated matrix.