Consider the set $$M := \{(x,y,z) \in \mathbb{R}^3 \mid x^2+y^2 = 1, (x-1)^2+y^2+z^2 = r^2 \}.$$
For which values of $r \in \mathbb{R}$ is $M$ a submanifold?
I think this exercise is supposed to be solved via the regular value theorem, which we described in the lecture as follows:
Regular Value Theorem: Assume that $M \subset \mathbb{R}^d$ is defined implicitly by $M = f^{-1}(a) \text{, where $f: U \subseteq \mathbb{R}^d \rightarrow \mathbb{R}^k$}. $ If $f$ is a $\mathcal{C}^r$ with regular value $a$ ($Df_x$ is onto for all $x \in M$), then $M$ is a $(d-k)$-dimensional submanifold of $\mathcal{C}^r$ smoothness.
Here is what I got so far:
I understand that we can describe $M$ implicitely via $M = f^{-1} \begin{pmatrix}1 \\ r \end{pmatrix}$ for
$$f(x,y,z) := \begin{pmatrix}x^2+y^2\\(x-1)^2+y^2+z^2\end{pmatrix}.$$
and we have
$$Df_x(v) = v_1 \begin{pmatrix}2x\\2x-2\end{pmatrix} + v_2 \begin{pmatrix}2y\\2y\end{pmatrix} + v_3 \begin{pmatrix}0\\2z\end{pmatrix} = \begin{pmatrix} 2x &2y &0 \\ 2x-2 &2y &2z \end{pmatrix}v$$
, so
$$ Df_x = 2\begin{pmatrix} x &y &0 \\ x-1 &y &z \end{pmatrix}.$$
We now need to consider the $(x,y,z)$ with $x^2+y^2 = 1$ and $(x-1)^2+y^2+z^2 = r^2$, for which $Df_x$ has rank $2$. I can see that for $r = 1$ $M$ is not a submanifold since $(1,0,0)$ fullfills the conditions but, $Df_{(1,0,0)}$ only has rank $1$.
Are there any other problematic values for $r$?