Suppose I have two circles, of radii $1$ and $R <<1$, respectively. Denote the boundary of the first circle of radius $1$ by $\gamma_{1}$. Denote the boundary of circle of radius $R$ by $\gamma_{2}$. I position $\gamma_{2}$ in the interior of $\gamma_{1}$, such that $\gamma_{2}$ is $\mathbf{tangent}$ to $\gamma_{1}$ at one point.
Now each circle individually is clearly an analytic curve (or generally, an analytic manifold). What about the curve $\gamma_{1} \cup \gamma_{2}$? Is it also analytic everywhere, or is it no longer differentiable at the point of tangency?
I am guessing it is indeed still analytic, if one uses the formal definition of a manifold with its fundamental characteristics ($\gamma_{1} \cup \gamma_{2}$ can be covered by a countable collection of open sets, such that there exists a $C^{\omega}$ diffeomorphism on each open set taking it to $\mathbb{R}$).
To make things very precise. Parametrise $\gamma_{1}$ by $\varphi$ as
$$\gamma_{1}(\varphi) = \{(\cos \varphi, \sin \varphi): \quad \varphi \in (-\pi, \pi) \}.$$
Parametrise $\gamma_{2}$ by $\phi$ as
$$\gamma_{1}(\varphi) = \{(-R\cos \phi - (1-R), R\sin \phi): \quad \phi \in (0, 2\pi) \}.$$
Geometrically on the $(x,y)$ plane this means that $\gamma_{1}$ centre is at the origin, while $\gamma_{2}$ centre is at $(-1+R, 0)$; $\gamma_{1}$ and $\gamma_{2}$ are tangent to each other at $(-1,0)$.
Is the resulting union of two circles analytic (at $(-1,0)$) ?
At the point $x \in \gamma_1 \cup \gamma_2$ where the circles $\gamma_1,\gamma_2$ are tangent, the following fact is true: for every sufficiently small open neighborhood $U \subset \mathbb{R}^2$ of $x$, the set $$\bigl(U \cap (\gamma_1 \cup \gamma_2)\bigr) - \{x\} $$ has at least four connected components: two open arcs on $\gamma_1$ incident to $x$, and two open arcs on $\gamma_2$ incident to $x$.
It follows that $\gamma_1 \cup \gamma_2$ is not even a topological $1$-manifold at the point $x$, let alone an analytic $1$-manifold: if it were a topological 1-manifold, then there would exist arbitrarily small open neighborhoods $U \subset \mathbb{R}^2$ of $x$ such that $U-x$ has exactly two connected components.