An alternative method for a question in Baby Rudin chapter 1 exercise 16 I'm trying to solve the following problem in Baby Rudin :

Suppose $k\ge 3$, $x,y \in \Bbb{R}^k,|x-y|=d>0$, and $r>0$ Prove:
(a) If $2r>d$, there are infinity many $z$  such that: $$ |z-x|=|z-y|=r $$

I solved the problem through constructing a unit vector perpendicular to $(x-y)$ and showing there are infinity many of these under the constraint. I'm trying to solve the problem by another way , I got a hint that it can be solved through constructing a system of equations which is under-determined here's my attempt (the problem I ran into is stated at the end):
$|z-x|^2=(z-x)(z-x)=|z|^2 -2xz+|x|^2$
similarly, $|z-y|=|z|^2-2yz-|y|^2$
Then $ |z-x|=|z-y|=r $ when $z(x-y)=\frac 1 2(|x|^2-|y|^2)$
Also since we want $ |z-x|=r$ then $|z-x|=|rw|$ where $w \in \Bbb{R}^k$ and $|w|=1$, this gives $z=x+rw$ combining this with $|y-z|^2=r^2$ gives $r^2=|x+rw-y|^2=d^2+2rw(x-y)+r^2\implies w  (y-x)=\displaystyle\frac {d^2} {2r}=\frac{d}{2r}|x-y| $
then since the left hand side is a dot product we get $$w_1(x_1+y_1)+w_2(x_2+y_2)+...+w_k(x_k+y_k)=\frac{d}{2r}|x-y| $$
and since $w$ is a unit vector we have $\sqrt{w_1^2+w_2^2+...w_k^2}=1$
Hence we have a system of two equations and $k$ unknowns and $k\ge3$ then the system has infinity many solutions since the number of $w$'s we  get is the same as the number of $z$'s.
What I'm not sure about is that the second equation is non-linear, does that matter?
 A: There is no real way around your initial solution.
Your second strategy essentially boils down to the same approach.
The problem is invariant by affine transformation.
So without loss of generality, we can assume $y=0$, $|x|=d>0$ and $r=1$.
The condition $|z-x|=|z|=r$ is equivalent to $|z-x|^2=|z|^2$ and $|z|=1$.
Hence your are looking for all $z$ such that
$$
(z,x)=\frac{d^2}{2}\quad\mbox{and}\quad |z|=1.
$$
ie
$$
x_1z_1+\ldots+x_kz_k=\frac{d^2}{2}\quad\mbox{and}\quad z_1^2+\ldots+z_k^2=1.
$$
Now $x/2$ is a solution of the lhs linear equation, so setting $z=x/2+h$, this amounts to solving
$$
x_1h_1+\ldots+x_kh_k=0\quad\mbox{and}\quad |x/2+h|^2=1.
$$
Observe that the lhs means that $x$ and $h$ are orthogonal, so $|x/2+h|^2=|x/2|^2+|h|^2= 1$ so this is now equivalent to
$$
x_1h_1+\ldots+x_kh_k=0\quad\mbox{and}\quad |h|^2=1-\frac{d^2}{4}.
$$
If $d>2$, there are clearly no solutions. If $d=2$, there is exactly one solution, $h=0$.
So assume $d<2$. The solutions of the lhs constitute a hyperplane, ie a $k-1$ dimensional subspace of $\mathbb{R}^k$.
So the solution set is a sphere in subspace of dimension $k-1\geq 2$.
This is infinite.
