Two circles whose points are equidistant I'm trying to solve this problem following ajotatxe's approach in a question I asked previously:

(a) In the space $\mathbb{R}^k$ a two-dimensional sphere $S^2$ and a circle $S^1$ are situated so that the distance from any point of the sphere to any point of the circle is the same.
  Is this possible?
(b) Consider problem a) for spheres $S^m$, $S^n$ of arbitrary dimension in $\mathbb{R}^k$. Under
  what relation on $m$ , $n$ , and $k$ is this situation possible?

Suppose a sphere $S^m \subseteq \mathbb{R}^k$ satisfies $\| x - x_1 \| = r_1$ and $x
= A y + b$ for orthonormal column vectors $A \in \mathbb{R}^{k \times m}$,
i.e. $A^T A = I_m$.
I first tried to prove that the projection of $x_1$ onto the affine subspace $y_1 = b + A A^T (x_1 - b)$ still has the same distance for all $x$, so the distance constraint is easier to handle. But I'm stuck below, is this claim true?
$\| x - y_1 \| = \| A y - A A^T (x_1 - b) \| = \| y - A^T (x_1
- b) \| = \| A^T (x - b) - A^T (x_1 - b) \| = \| A^T (x - x_1) \|$
 A: I'm assuming these are metric spheres corresponding to Euclidean distance, not just topological spheres.
(a) It works in $\mathbb R^5$.  Take the sphere $x_1^2 + x_2^2 + x_3^2 = 1$, $x_4 = x_5 = 0$, and the circle $x_4^2 + x_5^2 = 1$, $x_1 = x_2 = x_3 = 0$.
Similarly in higher dimensions.
EDIT
(b) Similarly if $k \ge m+n+2$.
Moreover  this is necessary as well as sufficient.  Let $a$ be the centre of the $m$-sphere and $b$ the centre of the $n$-sphere, $r_m$ and $r_n$ their radii.  Let $M$ and $N$ be the linear spans of the vectors $s - a$ for $s$ in the $m$-sphere and $t - b$ for $t$ in the $n$-sphere respectively. We need
$\|(x+a) - (y + b)\|$ to be constant for $x \in M$, $y \in N$ with $\|x\| = r_m$, $\|y\| = r_n$.  Now
$$ \|(x+a) - (y + b)\|^2 = \|x\|^2 + \|a - y - b\|^2 + 2 x \cdot (a - y - b)$$ 
 Since $x$ and $-x$ must give the same value, $x \cdot (a - y - b) = 0$ for all $x \in M$, i.e. $(a-b) + N \subseteq M^\perp$.
Now $M$ has dimension $m+1$, so $M^\perp$ has dimension $k-m-1$, and if this contains a translate of the $n+1$-dimensional space $N$ we must have $n+1 \le k-m-1$, i.e. $m+n+2 \le k$.
