Rigorously showing there are infinitely many points of intersection? I'm working on a problem that states if $k\geq 3$, $x,y\in\mathbb{R}^k$, $|x-y|=d>0$, and $r>0$, then

(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 $z$.

Intuitively and geometrically, this is obvious. The $z$ are all points distance $r$ from both $x$ and $y$, so they are the surfaces of the balls of radius $r$ centered around $x$ and $y$. If $2r>d$, then the radii are greater than half the distance between $x$ and $y$, and thus the balls will intersect in some surface, and if $k\geq 3$, then there will be infintely many points on this surface.
If $2r=d$, these balls will intersect in exactly one point of tangency.
Is there a way to more rigorously show these results without appealing to geometric intuition? For example, is there some algebraic system that you could exhibit in each case and say, here there are infinitely many solutions or there is exactly one solution, respectively? Thanks.
 A: For part (b), you get it from triangle equality that $z = \alpha x + (1-\alpha)y$, $\alpha \in [0,1]$. Hence, $$||z-x|| = (1-\alpha) || y-x||$$ which gives us $\alpha = \frac12$. Hence, $\displaystyle z = \frac{x+y}{2}$.
For part (a), try to parameterize the intersecting manifold. Note that $$z = \frac{x+y}{2} + v \sqrt{r^2 - \left(\frac{d}2\right)^2}$$ where $v$ is a unit vector perpendicular to $y-x$. Note that when $k \geq 3$, there are infinite unit vectors perpendicular to a given vector. Hence, there are infinitely many $z \in \mathbb{R}^k$.
A: Your geometric argument is perfectly rigorous. If you want it more "formal" you can argue as follows:
One may assume $x={d\over 2} e$, $y=-{d\over2} e$ for some unit vector $e$. The orthogonal complement $U$ of $\langle e\rangle$ has dimension $k-1$. Any point $z\in{\mathbb R}^k$ can be written as $z=\lambda e+ u$ for some $\lambda\in{\mathbb R}$ and some vector $u\in U$. The condition $|z-x|=|z-y|$ is equivalent to $|z-x|^2=|z-y|^2$, and this in turn is equivalent to
$$\bigl(\lambda -{d\over2}\bigr)^2 +|u|^2=\bigl(\lambda +{d\over2}\bigr)^2 +|u|^2$$
which is true iff $\lambda=0$ (as expected!).
When an $r>0$ is given and we have the additional condition $|z-x|=r$ then the admissible $z$'s are the vectors $u\in U$ with ${d^2\over 4}+|u|^2=r^2$ or
$$|u|^2={1\over 4}(4r^2-d^2)\ .$$
It follows that if $2r<d$ then there is no such $u$, if $2r=d$ then there is only the solution $u=0$, and if $2r>d$ then all points on the $(k-2)$-dimensional sphere $S\subset U$ of radius ${1\over2}\sqrt{4r^2-d^2}>0$ satisfy the given conditions. As $k-2\geq1$ there are infinitely many such points.
