Line segment connecting two concentric circles with a fixed midpoint Consider two circles $C_1, C_2$ centered at the origin $(0,0)$ in $\mathbb R^2$ with radius $r_1$ and $r_2$ respectively.
Question: Find all possible pairs $\{r_1,r_2\}$ such that there exists a line segment $\ell=\overline{p_1p_2}$ connecting two points $p_i\in C_i$ in each circle such that the midpoint of $\ell$ is $(1,0)$.
For example, some obvious solutions are as follows: $\{r_1,r_2\} =\{1-\epsilon, 1+\epsilon\}$ for some $0<\epsilon<1$ or $r_1=r_2$. But definitely there will be more solutions.
 A: For a circle $C_1$ of radius $r_1$, for every point $p$ in $C_1$ there is a corresponding radius $r_p$ which is simply the length $Op'$ where $p'$ is the reflection of $p$ in $(1, 0)$.
Let $p$ be given by the coordinates $(r_1\cos\theta, r_1\sin\theta)$. Then $p'$ is located at $(2 - r_1\cos\theta, -r_1\sin\theta)$, and the length $Op'$ is given by $\sqrt{4 - 4r_1\cos\theta + r_1^2}$
This value varies continuously on $\theta$, with minimums and maximums of $\sqrt{4 - 4r_1 + r_1^2}$ and $\sqrt{4 + 4r_1 + r_1^2}$ respectively. This means, if $r_1 \geq 2$ that $r_p$ lies in the interval $[r_1-2, r_1+2]$. That is, a solution $\{r_1, r_2\}$ is valid if $|r_1 - r_2| \leq 2$ and $r_1 \geq 2$
For $r_1 < 2$, the minimum becomes $2 - r_1$ instead, and so $r_p$ lies in the interval $[2-r_1, 2+r_1$].
So, the overall solution set can be represented, rather uglily, as
$$S  = \{(r_1, r_2) \in {\mathbb{R}^+}^2 | r_1 \geq 2, |r_1 - r_2| \leq 2\} \cup \{(r_1, r_2) \in {\mathbb{R}^+}^2 | r_1<2, |r_2-2|\leq r_1\}$$
