When is the projection of an ellipsoid a circle? Consider an ellipsoid in the three dimensional Euclidean space, say
$$\frac{x^2}{a^2}+\frac{y^2}{b^2} + \frac{z^2}{c^2} =1 $$
where $a$, $b$, $c$ are positive reals.
I'm counting the number of planes through the origin so that the image is a perfect circle. There may be divergent cases if we consider the case that some of $a$, $b$, $c$ are coincide. But at first, let us focus on the case that $a$, $b$, $c$ are all different, say $a>b>c$.
I guess the answer would be $4$. I have made many efforts but failed.
What I have observed is the that at least two such planes exists and the radius of the circle is $b$. Just consider rotating plane possesses $y$ axis and apply intermediate value theorem.
Causion! We are concerning projection, not intersection.
PS. Now I guess there are infinitely many...
PS2. According to one suggested answer, there are just two such planes for the non-degenerate case. I'm checking if it is correct.
PS3. Another opinion appeared that the selected answer may have fault. And it seems making sense. I think somewhat stronger analysis is required.
PS4. The above PS3 is about another answer which now have disappeared.

Considering an enveloping cylinder, namely
$$
\left(
  \frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}-1
\right)
\left(
  \frac{l^2}{a^2}+\frac{m^2}{b^2}+\frac{n^2}{c^2}
\right)=
\left(
  \frac{l x}{a^2}+\frac{m y}{b^2}+\frac{nz}{c^2}
\right)^2$$
The eigenvalues $\lambda$ are given by
$$0=\det
\begin{pmatrix}
   \frac{1}{a^2} \left( \frac{m^2}{b^2}+\frac{n^2}{c^2} \right)-\lambda &
  -\frac{l m}{a^2 b^2} & -\frac{l n}{a^2 c^2} \\
  -\frac{m l}{b^2 a^2} &
   \frac{1}{b^2} \left( \frac{n^2}{c^2}+\frac{l^2}{a^2} \right)-\lambda &
  -\frac{m n}{b^2 c^2} \\
  -\frac{n l}{c^2 a^2} & -\frac{n m}{c^2 b^2} &
   \frac{1}{c^2} \left( \frac{l^2}{a^2}+\frac{m^2}{b^2} \right)-\lambda \\
\end{pmatrix}$$
where $\lambda=0$ is one of the roots.
The discriminant (with last factor in cyclic permutations of $a$, $b$ and $c$) is given by
$$\Delta
=\frac{(l^2+m^2+n^2)^2}{a^8 b^8 c^8}
\left( \frac{l^2}{a^2}+\frac{m^2}{b^2}+\frac{n^2}{c^2} \right)
[(\alpha l^2-\gamma n^2)^2-
  2\beta m^2 (\alpha l^2+\gamma n^2)+
  \beta^2 m^4
]$$
where $\alpha=b^2-c^2$, $\beta=c^2-a^2$, $\gamma=a^2-b^2$.
Note that for $a>b>c>0$,
$$\fbox{$\alpha,-\beta,\gamma>0$}$$
For circular cylinder, there are two equal eigenvalues implying  $\Delta=0$, hence
$$\fbox{$\frac{l^2}{n^2}=\frac{\gamma}{\alpha} \land m=0$}$$
The enveloping circular cylinder is given by
$$
\fbox{$\frac{b^2}{a^2 c^2}
\left(
  \frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}-1
\right)=
\left(
  \sqrt{\frac{a^2-b^2}{a^2-c^2}} \, \frac{x}{a^2} \pm
  \sqrt{\frac{b^2-c^2}{a^2-c^2}} \, \frac{z}{c^2}
\right)^2
$}$$
Further points to be noticed

*

*The touching conic is not circular and lying on the plane

$$\color{green}{z=\pm \frac{c^2}{a^2} \sqrt{\frac{a^2-b^2}{b^2-c^2}}\, x}$$

*

*The uniform cross section of the cylinder lies on the plane

$$\color{blue}{z=\pm \sqrt{\frac{a^2-b^2}{b^2-c^2}}\, x}$$

*

*The circular section for the ellipsoid lies on the plane

$$\color{red}{z=\pm \frac{c}{a} \sqrt{\frac{a^2-b^2}{b^2-c^2}}\, x}$$

