TITLE: Do 2 of the same circles contain each other? (Proving / Disproving)
QUESTION [0 upvotes]: Two Circles, $C_1$ and $C_2$ have the same center $(h, k)$ and radius $r$.
Trying to prove or disprove that $C_1$ contains $C_2$ and $C_2$ contains $C_1$ OR neither contains the other.. but I think I'm leaning towards the former by this logic:
A Circle $A$ is contained in a circle $B \iff \forall \; (x, y) \in A$, $(x, y)$ is also $\in B$.
Or in other words, every point in B is less than A's radius from A's Center.
So for the case where $A$ and $B$, are geometrically identical:
The distance from $A$'s center to $B$'s center is $0$, therefore any point in $B$ is at most $A$'s radius away from $B$'s center (aka $B$'s radius, so within $B$).
Another approach might hint at the Ship of Theseus. Since $A$ and $B$ are geometrically identical, $B$ can be replaced with $A$. and $A$ contains itself. Therefore $A$ contains $B$, and $B$ contains $A$ by the same logic.
Does this hold water? I started by thinking about counting intersections but that sent me down the wrong rabbit hole I think.

REPLY [0 votes]: Ultimately depends on how formal you want to be, but your first approach feels vague, and the second feels circular (in that it somewhat assumes $A=B$ from the outset).
I would approach it in a more set-theoretic fashion.
You seem to have defined two sets, your circles, by the following rules:
$$\begin{align*}
S_A &:= \left\{ (x,y) \in \mathbb{R}^2 \,\middle|\, 0 \le \sqrt{|x-h|^2 + |y-k|^2} < r  \right\} \\
S_B &:= \left\{ (x,y) \in \mathbb{R}^2 \,\middle|\, 0 \le \sqrt{|x-h|^2 + |y-k|^2} < r  \right\} 
\end{align*}$$
Notice how both have encoded that, for $(x,y)$ to be contained in the circle, its distance from $(h,k)$ must be less than $r$. (Possibly including $r$ depending on your definition, but it won't affect things.)
Then that $S_A \subseteq S_B$ and $S_B \subseteq S_A$ is immediate ($(x,y) \in S_A$ iff it meets the distance condition, which is the same in $S_B$), and so $S_A = S_B$.
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