When the intersection between a sphere and a cylinder is planar?

Question

We have a sphere and a circular cylinder. Let the sphere center be $O$ and radius $R$, and the cylinder axis $a$ and radius $r$. I solved the specific case intersection graphically on 2 planar projections, but I would like to have a general understanding classification of cases, hence this question. So is this correct?

1. If a cylinder has its axis through the sphere's center, $O\in a$, the intersection is planar (a circle), if the intersection exists ($r\le R$). Here in both projections $O_1\in a_1$ and $O_2\in a_2$.
2. If the cylinder's axis in one projection (e.g. plan) passes through the projection of the center but not in the other (e.g. front), or $O_1\in a_1$ but $O_2\notin a_2$ , or in other words the cylinder position is "through the sphere middle" but not through the center, here I'm not sure. E.g. in my specific example the cylinder is horizontal, $r<R$, and $a$ is directly below $O$. In $projection_1$ the intersection is a line (looks planar), in $projection_2$ it is an ellipse. Which looks like a projection of a circle. However, if the section is planar, then a diagonal section of a cylinder should be an ellipse. But a planar section of a sphere is a circle. Could it be that it just looks planar due to inaccuracy of construction? I have similar looking result in another example too.
3. If $O_1\notin a_1$ and $O_2\notin a_2$, then intersection is two non-planar "rings"
4. A special case when cylinder touches the sphere, it looks like a 3d figure-8

Answer 1

It is way easier to think when a plane section of a cylinder, that is an ellipse having ratio between the axis that depends on the angle between the cutting plane and the axis of the cylinder, is also a plane section for a sphere i.e. a circle. There is no inaccuracy in point $(2)$: in order to have a planar intersection, the center of the sphere has to lie on the axis on the cylinder.

Answer 2

We may assume that the sphere is given by $$S:\quad x^2+y^2+z^2=a^2, \qquad a>0\ ,$$ and the cylinder by $$C:\quad (x-m)^2 +y^2=b^2,\qquad m\geq0, \quad b>0\ .$$ The sphere intersects the $(x,y)$-plane in the circle $K_s:\ x^2+y^2=a^2$, and the cylinder in the circle $K_c: \ (x-m)^2+y^2=b^2$. The vertical projection of the curve $\gamma:=S\cap C$ is a subset of $K_c$.

We now have to distinguish cases according to the way the two circles $K_s$ and $K_c$ intersect. The two circles (i) may not intersect, they (ii) may touch each other in one point, they (iii) may intersect in two points, or they (iv) may coincide.

(i) If $K_c$ lies in the exterior of $K_s$, or if $K_c$ contains $K_s$ in its interior, then $\gamma=S\cap C=\emptyset$.

If If $K_c$ lies in the interior of $K_s$ then $\gamma$ consists of two separated loops. These loops are not planar unless $m=0$, on account of the logic you have put forward in your question.

(ii) If $K_c$ touches $K_s$ in one point and does not intersect the interior of $K_s$ then $\gamma$ degenerates to this point. When $K_c$ touches $K_s$ in the point $A:=(a,0,0)$ from the inside then $b<a$, and $\gamma$ is a figure eight with singular point $A$. The direction of the tangents of $\gamma$ at $A$ depend on the values of $a$ and $b$.

(iii) When $K_c$ and $K_s$ intersect in two points $P_1$ and $P_2$ lying symmetrically with respect to the $x$-axis the curve $\gamma$ is a single nonplanar loop crossing the $(x,y)$-plane vertically in the two points $P_1$ and $P_2$.

(iv) When $K_c=K_s$ then $\gamma=K_c=K_s$.