Line segment connecting two concentric circles with a fixed midpoint

Question

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.

Answer 1

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

$$S = \{(r_1, r_2) \in {\mathbb{R}^+}^2 | r_1 \geq 2, |r_1 - r_2| \leq 2\} \cup \{(r_1, r_2) \in {\mathbb{R}^+}^2 | r_1<2, |r_2-2|\leq r_1\}$$

Answer 2

Let $p_1$ be on circle $C_1$ and $p_2$ be on circle $C_2$, then

$ p_1^T p_1 = r_1^2 $, and $ p_2^T p_2 = r_2^2 $

And it is required that $ p_1 + p_2 = (2, 0) \hspace{50pt} (1)$

By writing $p_1 = r_1 (\cos(\theta_1) ,\sin(\theta_1)), p_2 = r_2 (\cos(\theta_2), \sin(\theta_2))$, this is translated into

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$ r_1 \cos(\theta_1) + r_2 \cos(\theta_2) = 2 \hspace{50pt} (2)$

$ r_1 \sin(\theta_1) + r_2 \sin(\theta_2) = 0\hspace{50pt} (3) $

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If we define

$u = [\cos(\theta_1), \sin(\theta_1) ]^T $

and

$v = [\cos(\theta_2), \sin(\theta_2) ]^T$

then the above two equations can be written in matrix-vector format as

$ A u + B v = w \hspace{50pt} (4)$

where

$ A = r_1 I, B = r_2 I, w = [2, 0]^T\hspace{50pt} (5) $

Solving for $v$

$ v = B^{-1} ( w - A u) = \dfrac{1}{r_2} (w - r_1 u) \hspace{50pt} (6) $

But $v$ is a unit vector, i.e. $v^T v = 1 $, hence

$ (w - r_1 u)^T (w - r_1 u) = r_2^2 \hspace{50pt} (7)$

This simplifies to

$ w^T w - 2 r_1 w^T u + r_1^2 = r_2^2 \hspace{50pt} (8)$

Substituting $w$, $(8)$ becomes

$$ 4 r_1 \cos(\theta_1) = 4 + r_1^2 - r_2^2 \hspace{50pt}(8a) $$

Once you have $u$ (and there can be zero, one or two solutions), one can evaluate the corresponding $v$, from equation (6)$.

Equation $(8)$ gives a necessary condition for the solvability of this problem which is

$ 4 r_1 \ge | 4 + r_1^2 - r_2^2 | \hspace{50pt} (9) $

If we were to solve for $u$ instead of $v$ from equation $(4)$, then we would end up with following necessary condition for the solvability of the problem,

$ 4 r_2 \ge | r_1^2 - r_2^2 - 4 | \hspace{50pt} (10)$

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Suppose $r_1 = 1$ , $r_2 = 3$ , then using equation $(8a)$

$4 \cos(\theta_1) = -4 $

which gives one solution, $\theta_1 = \pi $, and using equation $(6)$ we get

$ v = \dfrac{1}{3} ( (2, 0) - (-1, 0) ) = (1, 0) $

Therefore, $\theta_2 = 0 $

So that the line segment is $p_1 p_2$ with $p_1 = (-1, 0) , p_2 = (3, 0) $

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As another example, Suppose $r_1 = 5$ and $r_2 = 6$ , then

$ 20 \cos(\theta_1) = -7 $

which has two solutions

$ \theta_{1a} = 1.9283674 , and $ \theta_{1b} = 4.3548178 $

Using the first of these two, in equation $(6)$ gives

$ v = \dfrac{1}{6} ( (2, 0) - 5 (-0.35, 0.9367497 ) ) \\ = ( 0.625, -0.78062475)$

From which $\theta_{2a} = 5.3875205 $

And using the second solution of $\theta_1$ which is $\theta_{1b} = 4.3548178 $ in equationi $(6)$ gives

$ v = \dfrac{1}{6} ( (2, 0) - 5 (-0.35, -0.9367497 ) ) = (0.625, 0.78062475) $

From which $\theta_{2b} = 0.895665 $

Therefore, the two possible line segments $p_1 p_2 $ (with p_1 lying on the circle with radius $5$ and $p_2$ lying on the circle of radius $6$ ) are

LINE SEGMENT 1: $ (-1.75, -4.683748) - (3.75, 4.683748) $

LINE SEGMENT 2 : $ (-1.75, 4.683748) - (3.75 , -4.68375) $

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