I want to prove that for any circle $C$ with $r \gt 0$ and center $(x_0,y_0)$ in the Euclidean Plane $\mathbb R \times \mathbb R$, if $(x_1,y_1), (x_2,y_2) \in C$ and are separated by $180\deg$, then the line $L$ that connects $(x_1,y_1)$ to $(x_2,y_2)$ must pass through $(x_0,y_0)$. In particular, then, this line must contain the diameter of $C$.

If there is a more straightforward non-geometric approach, I would love to see it.

Firstly, what does it mean for $(x_1,y_1)$ to be separated by $(x_2,y_2)$ by $180 \deg$? Consider the Euclidean distance $d$ between $(x_1,y_1)$ and the center of the circle. Next, consider the angle $\theta_1$ between the positive $x$ axis and the radial line connecting $(x_0,y_0)$ to $(x_1,y_1)$. Let $\theta_2=\theta_1+\pi$ and let this correspond to the angle between the positive $x$ axis and the radial line connecting $(x_0,y_0)$ to $(x_2,y_2)$. Because $(x_2,y_2) \in C$, we can rewrite $(x_2,y_2)$ by making the following observations: \begin{align}&(1) \quad x_2=x_0+\cos(\theta_2)d \implies x_2=x_0+\cos(\theta_1+\pi) \implies x_2=x_0-\cos(\theta_1)\\&(2) \quad y_2=y_0+\sin(\theta_2)d \implies y_2=y_0+\sin(\theta_1+\pi) \implies y_2=y_0-\sin(\theta_1)\end{align}

By definition of $C$, $\sin(\theta_1)=y_1-y_0$ and $\cos(\theta_1)=x_1-x_0$.

Therefore, $(1)$ and $(2)$ can be rewritten as:

\begin{align} &(3) \quad x_2=x_0-(x_1-x_0) \implies x_2=2x_0-x_1 \\&(4) \quad y_1=y_0-(y_1-y_0) \implies y_2=2y_0-y_1\end{align}

As such, $(x_2,y_2)=(2x_0-x_1,2y_0-y_1)$.

Next, consider a line represented by the following equation: $L(x)=\frac{y_2-y_1}{x_2-x_1}(x-x_1)+y_1$. Plugging in our values for $x_2$ and $y_2$, we have:

$$L(x)=\frac{2y_0-y_1-y_1}{2x_0-x_1-x_1}(x-x_1)+y_1 \implies L(x)=\frac{y_0-y_1}{x_0-x_1}(x-x_1)+y_1$$

Now, solving for when $x=x_0$, we have that $L(x_0)=y_0$, which means that $(x_0,y_0) \in L$.

This argument worked for any arbitrary ordered pairs meeting our initial criteria, as well as for any circle in the Euclidean plane.

  • $\begingroup$ Call the points $P$ and $Q$, the center, $R$. Saying $P$ and $Q$ are separated by $180^{\circ}$ is saying the angle formed at $R$ by the segments $PR$ and $QR$ is $180^{\circ}$. But that means $R$ is on the line joining $P$ and $Q$. $\endgroup$ Mar 21, 2022 at 6:47

2 Answers 2


Whether there is a more straightforward approach depends on how you define "circle" and "separation by 180 degrees".

I can define a circle $S(a,r)\subset\mathbb{C}$ to be the loci of points $z$ such that $|z-a|=r$ for $a\in\mathbb{C}$ and $r\in\mathbb{R}^+$.

I can then define $z,z'\in S(a,r)$ to be separated by 180 degrees whenever $z-a=e^{\pi i}(z'-a)$, since 180 degrees is equal to $\pi$ radians. But $e^{\pi i}=-1$, so the condition amounts to $z-a=a-z'$.

The separation of $z$ and $z'$ by 180 degrees implies that $z\neq z'$. How? Suppose for contradiction that $z=z'$. Then $2z=2a$, which is not allowed by $|z-a|=r>0$.

The line passing through $z$ and $z'$ is defined as the set of points $z+(z'-z)t$, where $t$ ranges over $\mathbb{R}$.

Observe that $z+\dfrac{1}{2}(z'-z)=\dfrac{1}{2}(z+z')=\dfrac{1}{2}\cdot2a=a$.

  • 1
    $\begingroup$ Very cool. I'm not as experienced with complex numbers but this is much more attractive in flavor than mine. Thanks! $\endgroup$ Mar 21, 2022 at 6:35

This is a proof (outline, left for the reader to fill in the details) by reducing to a special case (and is not the most simplified case, but it is sufficient). I prefer this proof to the nicely presented proof by Chris Sanders using complex numbers because it generalizes to higher dimensions more easily (i.e., we can replace $\mathbb{R}^2$ with $\mathbb{R}^n$ and nothing changes). Before we get into the proof, some definitions:

Definition 1 (Anti-parallel):

Two vectors $x,y\in\mathbb{R}^2$ are said to be anti-parallel or separated by $180^{\circ}$ provided that $\langle x,y\rangle =-\lVert x\rVert\cdot\lVert y\rVert$.

Definition 2 (Circle in the plane):

For $r\in(0,\infty)$ and $z_0\in\mathbb{R}^2$, define $S_r(z_0)=\{z\in\mathbb{R}^2:\lVert z - z_0\rVert = r\}$ to be the circle of radius $r$ centered at $z_0$.

Proof (of main result):

It suffices to show that the result holds for the circle $S_r(0)$ and any two points $x,y\in S_r(0)$ that are anti-parallel. I encourage you to try and show that the line connecting them, parametrized by the function $\ell:\mathbb{R}\to\mathbb{R}^2$ $$\ell(t)=x+t(y-x)$$ passes through the origin (this relies on the fact that $\lVert x\rVert = \lVert y\rVert = r$).

The reason why this case is enough is because for arbitrary $z_0$, we can consider the image of $S_r(z_0)$ under the translation $T:\mathbb{R}^2\to\mathbb{R}^2$ given by $T(z) = z-z_0$ along with considering its inverse, $T^{-1}(z) = z + z_0$.

You should show that if $x,y\in S_r(z_0)$ and are separated by $180^{\circ}$ then $T(x),T(y)\in S_r(0)$ and $T(x),T(y)$ are still separated by $180^{\circ}$. Then use the result for $S_r(0)$ along with $T^{-1}$ to show that the line adjoining $x$ and $y$ passes through $z_0$.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .