<s>[INST] 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.
 [/INST] For convenience, we use $$ax^2+by^2+cz^2=1.$$
Consider the parametric lines of constant direction $(u,v,w)$ through some point $(x,y,z)$ on the ellipsoid.
$$a(x+tu)^2+b(y+tv)^2+c(z+tw)^2=1$$
or, after simplification,
$$(au^2+bv^2+cw^2)t^2+2(aux+bvy+cwz)t=0.$$
We will obtain the silhouette line on the ellipsoid by expressing that the rays are tangent, i.e. that the $t$ equation has a double root at $t=0$, which simply gives us the plane
$$aux+bvy+cwz=0.$$
That plane intersects the ellipsoid along an ellipse and one can always find an oblique projection that makes it a circle !
If you mean an orthogonal projection, then we must find the plane(s) that intersect along a circle. We can do this by rotating the ellipsoid so that the plane goes to $Z=0$.
We use the rotation relations
$$\begin{pmatrix}x\\y\\z\end{pmatrix}=
\begin{pmatrix}0&\frac{B^2+C^2}{\sqrt{(B^2+C^2)^2+(AB)^2+(AC)^2}}&\frac{A}{\sqrt{A^2+B^2+C^2}}\\\frac{C}{\sqrt{B^2+C^2}}&\frac{-AB}{\sqrt{(B^2+C^2)^2+(AB)^2+(AC)^2}}&\frac{B}{\sqrt{A^2+B^2+C^2}}\\\frac{-B}{\sqrt{B^2+C^2}}&\frac{-AC}{\sqrt{(B^2+C^2)^2+(AB)^2+(AC)^2}}&\frac{C}{\sqrt{A^2+B^2+C^2}}\end{pmatrix}
\begin{pmatrix}X\\Y\\Z\end{pmatrix}.$$
When plugged in the ellipsoid equation, this gives the equation of a conic in the $XY$ plane. This conic is a circle when the quadratic coefficients are equal and the is no mixed term. This is more conveniently done by rewriting the matrix in spherical coordinates.
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