Is the intersection between two $n$-spheres an $(n-1)$-sphere?

Question

Is it true that the intersection between two $n$-sphere in $\mathbb{R}^n$ is a $(n-1)$-sphere if is not empty or a single point? I have tried to prove it but my only idea is to work with equations and apparently is not a good idea.

Answer 1

We pick one center at the origin for sphere $S_0$ (radius $a$) and the other at the point $(1,0,0,...,0)$ (radius $b$) for the sphere $S_1,$ so that the two equations are respectively $$x_1^2+x_2^2+\cdots + x_n^2=a^2,\\ (x_1-1)^2+x_2^2+ \cdots +x_n^2=b^2.$$ Subtracting these gives $2x_1-1=a^2-b^2,$ so that any point in the intersection $S_1 \cap S_2$ must lie in the hyperplane $x_1=c,$ where $c=(a^2+b^2+1)/2.$ In this hyperplane, one views its "origin" as the point $P=(c,x_2, \cdots x_n).$ Then the intersection, lying in $x_1=c,$ is a sphere centered at $P$ in that plane, and with equation $$x_2^2+x_3^2+\cdots + x_n^2=r^2,$$ where $r^2$ may be found from either equation for $S_1$ or for $S_2.$ Hopefully these match, namely $r^2=a^2-c^2=b^2-(c-1)^2.$ These do indeed match (which one would expect by symmetry) and the expression for $r^2$ may be put in the form $$r^2=\frac14[(a+b)^2-1][1-(a-b)^2].$$ We are viewing $a,b>0$ and if the first factor here is negative it corresponds to $|a+b|<1$ so the two radii don't add up to enough for the spheres to meet. (They are outside each other.) In cases I've tried, if the first factor is positive and the second is negative, it corresponds to the case where one sphere is completely inside the other.

For generally located centers $C_1,C_2$ I think one can rescale so the distance between $C_1$ and $C_2$ is one, and apply the preceding, at least to get the center and radius. This means using $C_1+t(C_2-C_1),$ in vector notation, where $t$ is obtained from the previous case (the formula for that $c$ and its radius should be obtained using the rescaled versions of the initial radii for the spheres centered at $C_1$ and $C_2.$ And in this general case the resulting sphere lies in a hyperplane perpendicular to the line joining the two centers, so may not be so easy to visualize.

Answer 2

Let us consider two spheres one centered at $0\in\mathbb{R}^n$, the other at $a\in \mathbb{R}^n$. $$\begin{aligned} S_0 = \mathbb{S}^{n-1}(0, r_0) &= \{x\in \mathbb{R}^n : \|x\|=r_0\} \\ S_a = \mathbb{S}^{n-1}(a, r_a) &= \{x\in \mathbb{R}^n : \|x-a\| = r_a\} \end{aligned}$$ In the following we assume without loss of generality $a\neq 0$

If $a=0$, then we only have a non-trivial intersection for $r_0=r_a$ and in that case it is the entire sphere

Idea

We can write a point in the intersection $y\in S_0 \cap S_a$ as $$ y = \lambda a + u $$ for some $u\in U:=a^\perp$ in the orthogonal space of $a$ and some $\lambda \in \mathbb{R}$. If you draw down two intersecting circles, then you will find that $\lambda a$ is going to be the center of our $(n-2)$-sphere. We need to show that only one $\lambda$ is valid to see that the intersection lies in the subspace $$ S_0\cap S_a \subseteq U+ \lambda a, $$ and we need to prove that $\|u\|=\mu$ for some constant $\mu$ to obtain $$ S_0 \cap S_a = \mathbb{S}^{n-2}(\lambda a, \mu). $$ If we can deduce a specific $\lambda$ and $\mu$ from the fact $y\in S_0\cap S_a$, then we have "$\subseteq$" with those parameters.

Finding $\lambda$ and $\mu$

Our assumption results in

1. Using $y\in S_0$ implies $$ r_0^2 = \|y\|^2 = \lambda^2 \|a\|^2 + \|u\|^2 $$
2. Using $y\in S_a$ implies $$\begin{aligned} r_a^2 &= \|y-a\|^2 = \|a\|^2 + \|y\|^2 -2\underbrace{\langle a, y\rangle}_{=\lambda \|a\|} \\ &=(1-2\lambda + \lambda^2) \|a\|^2 + \|u\|^2 \end{aligned}$$

If we solve for $\|u\|^2$ in both equations we get $$\tag{1} r_0^2 - \lambda^2 \|a\|^2 = \|u\|^2 = r_a^2-(1-2\lambda + \lambda^2)\|a\|^2 $$ Adding $\lambda^2\|a\|^2$ to both sides and reordering implies $$ r_0^2-r_a^2 + \|a\|^2 = 2\lambda \|a\|^2 $$ by $a\neq 0$ we have $$ \lambda = \frac12 + \frac{r_0^2 -r_a^2}{2\|a\|^2}. $$ So only this specific $\lambda$ is allowed, and we have found our center $\lambda a$. To find the radius of our new sphere, we can plug our $\lambda$ into one of the equations we found for $\|u\|^2$ i.e. $(1)$.

Since all of our transformations are reversible it is easy to see that "$\supseteq$" also holds (alternatively take a point from the $(n-2)$-sphere $\mathbb{S}^{n-2}(\lambda a,\mu)$ and test membership of $S_0$ and $S_a$).

Answer 3

Alternatively, here is a geometric proof with no algebra.

Let the spheres have centers $A$ and $B$ and radii $a$,$b$. Assuming the intersection is nonempty, consider any point $X$ on the intersection of the spheres. Let $O$ be the orthogonal projection of $X$ onto the line $\overleftrightarrow{AB}$. Let $S$ be the set

$S = \{Y \,|\, d(Y, O) = d(X,O) \text{ and } \overrightarrow{OY} \perp \overrightarrow{OA}$}

We claim $S$ is the intersection. Note that the $S$ is an $(n-1)$-sphere, since it is the set of points equidistant to $O$ that lie in the hyperplane orthogonal to $\overrightarrow{OX}$ (and passing through $X$.)

Let $Y$ be any other point on the intersection. Then $d(X,A) = d(Y,A) = a$ and $d(Y,B) = d(X, B) = b$. So by SSS, $\triangle AXB$ and $\triangle AYB$ are congruent. Then by AAS, say, $\triangle AXO$ and $\triangle AYO$ are congruent. It follows that $d(X,O) = d(Y,O)$. Also by congruence, $m \angle AOY = m \angle AOX$ which is $90^\circ$ by construction. So $Y \in S$.

Conversely, suppose $Y \in S$. Then by SAS, $\triangle AXO$ and $\triangle AYO$ are congruent, and so $d(A,X) = d(A,Y)$; and by the same argument $d(B,X) = d(B,Y)$, so that $Y$ lies on both spheres.