# Prove that for any circle in R^2, if two ordered pairs on the circle are separated by 180 degrees, then the connecting line passes through the center

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.

• 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 $Q$. Commented Mar 21, 2022 at 6:47

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

• Very cool. I'm not as experienced with complex numbers but this is much more attractive in flavor than mine. Thanks! Commented 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 $$z_0$$.