What is the intersection of a plane and a sphere? Is it necessarily a circle? Can it be an ellipse?

Question

A First Course in Complex Analysis by Matthias Beck, Gerald Marchesi, Dennis Pixton, and Lucas Sabalka Exer 3.27,Exer 3.28

(Exer 3.27) Consider the plane $H$ determined by the equation $x + y -z = 0$. What is a unit normal vector to $H$? Compute the image of $E:=H\cap \mathbb S^{2}$ under the stereographic projection $\Phi$.

Asked here and I answered agreeing with OP that the image is the circle $T: = \{|z-(1+i)|^2 = 3\}$, but now I ask:

Q1. Is $(2)$ below wrong?

$$\forall \ \text{planes} \ J, \exists \text{circle} \ C \in \mathbb S^2 : C = J \cap \mathbb S^2 \tag{2}$$

To clarify, compare that (2) is different from (1) below. We know that $$\forall \ \text{circles} \ C \in \mathbb S^2, \exists \ \text{plane} \ J : C = J \cap \mathbb S^2. \tag{1}$$

Q2. Do I go wrong in attempting to disprove (2) as follows?

I think $(2)$ contradicts Exer 3.27: I observe that $E$, whose image is the circle $T$, is both

• the intersection (WA) of a plane and $\mathbb S^2$ and yet

• an ellipse (WA) --> Or is it? Sphere-plane intersection seems to be circle (Wiki).

If $(2)$ is true, then for $J=H, \exists C \in \mathbb S^2: C = H \cap \mathbb S^2 =: E ↯$.

Q3. How is Exer 3.27 consistent with the succeeding Exer 3.28?

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(Exer 3.28) Prove that every circle in $\hat{\mathbb C}$ is the image of some circle in $\mathbb S^2$ under stereographic projection $\Phi$.

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I rephrase: $$\forall \ R \in \hat{\mathbb C}, \exists C \in \mathbb S^2 : \Phi(C)=R$$

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Now consider $T$, the circle in Exer 3.27 s.t. $\Phi(E)=T$.

It seems that by Exer 3.28, for $R= T, \exists C \in \mathbb S^2 : \Phi(C)=T$.

$$\therefore, \Phi(E) = \Phi(C) \ \text{while} \ E \ne C \ \text{, but} \ \Phi \ \text{is bijective ↯ ?}$$

Answer 1

As already verified, all intersections of a circle and plane are circles either geodesic/great or small. I avoid symbols in the answer here.

If 2-parameter surface normal coincides with 1-parameter circle normal then the intersection is a great circle. The intersection is a geodesic.

From differential geometry standpoint $\psi$ is between arc and meridian. Cylindrical coordinates $(r,z)$ Clairaut constant derivative $\dfrac{d(r \sin \psi)}{dz}=0$ forms with respect to an arbitrary North-South polar axis.

If the 2-parameter surface normal does not coincide with 1-parameter circle normal, but makes an angle $\gamma$ of relative latitude then the intersection is a small circle. Intersection is like a parallel circle but not an equator.

Clairaut constant derivative $\dfrac{d(r \sin \psi)}{dz}$ is a constant $ =\tan \gamma$ forms a small circle with respect to an arbitrary North-South polar axis.

In stereographic projection all non-equatorial sections of a plane passing through North pole of the sphere at angle $\gamma$ to North-South polar plane are small circles.

Answer 2

(Q1) I suspect some relationship is being illustrated. The 1st 2 coordinates in $[1,1,-1]$ are the centre of the circle $\Phi(E)$ while the norm of $[1,1,-1]$ s the radius of $\Phi(E)$.

(Q2) Yes if the planes intersect $\mathbb S^2$ or $\emptyset$ is considered a circle in $\mathbb S^2$.

(Q3), (Q4) E is not an ellipse. It is a circle in 3D (see here too) parametrised as (WA)

$$\begin{bmatrix} x_1(t)\\ x_2(t)\\ x_3(t) \end{bmatrix} = \begin{bmatrix} \sqrt{\frac 2 3} \cos[t]\\ -\sqrt{\frac 2 4} \sin[t] - \sqrt{\frac 2 {12}} \cos[t]\\ -\sqrt{\frac 2 4} \sin[t] + \sqrt{\frac 2 {12}} \cos[t] \end{bmatrix} = \begin{bmatrix} \sqrt{\frac 1 3}\\ -\sqrt{\frac 1 {12}}\\ \sqrt{\frac 1 {12}} \end{bmatrix}\sqrt{2}\cos(t)+ \begin{bmatrix} 0\\ -\sqrt{\frac 1 {4}}\\ -\sqrt{\frac 1 {4}} \end{bmatrix}\sqrt{2}\sin(t)$$

Observe that while $x^2+y^2+xy=\frac12$ is an ellipse, $x^2+y^2+xy=\frac12 \wedge z= x+y$ is a circle.