Taken from Rudin's Principles of Mathematical Analysis, 3rd edition (Chapter 1, Exercise 16):
Suppose $k\geq 3, x,y\in\mathbb{R}^k, |x-y|=d>0$, and $r>0$. Prove:
(a) If $2r > d$, there are infinitely many $z\in\mathbb{R}^k$ such that $|z-x|=|z-y|=r$
(b) If $2r=d$ there is exactly one such $z$.
(c) If $2r < d$, there is no such $z$.
Geometrically (at least in $\mathbb{R}^3$), if $2r>d$ then there exists a circumference of points $z$ such that $|z-x|=|z-y|=r$. I guess that for $\mathbb{R}^k$ the set of such points forms a $(k-2)$-sphere.
More formally, let $m$ be the midpoint between $x$ and $y$: $m=\frac{x+y}{2}$. Then we define the plane $\Pi$ as the set of $w\in\mathbb{R}^k$ that satisfy $(\lambda(x-y))w = (\lambda'(x-y))m$ for some $\lambda, \lambda'\in\mathbb{R}$. Let $S = \{p\in\Pi : |p-x|=r\}$. Because of its geometric interpretation it's obvious that $|p-x|=|p-y|$ for every $p\in S$, but I can't figure out how to prove it formally. $S$ can also be defined as the intersection of $\Pi$ and a $(k-1)$-sphere centered at $m$ with radius $\sqrt{r^2-(d/2)^2}$.
(b) is obvious since $m$ is the only possible point that fulfills what's asked.
(c) follows from the fact that, if we set $S$ as the $(k-1)$-sphere centered at $m$ with radius $\sqrt{r^2-(d/2)^2}$, its radius is non-existent, therefore no such sphere exists.
Is this approach correct? How can I prove (a)? Thanks in advance. (c)