The proof is a little long, but I think it is correct. If there are simpler methods, please share!

Let a circle $C$ be a set of ordered pairs from $\mathbb R \times \mathbb R$ satisfying the following equation: $(x-x_0)^2+(y-y_0)^2=r_0^2$, where $x_0, y_0,$ and $r_0 \gt 0$ are constants. Find the collection of pairwise ordered pairs $\left((x_1,y_1),(x_2,y_2) \right)$, where $(x_1,y_1) \in C$ and $(x_2,y_2) \in C$, that have the largest Euclidean Distance between them for all pairwise ordered pairs in $C \times C$ (not to be confused with $\mathbb C \times \mathbb C$). Subsequently show that the line segment between any such $(x_1,y_1)$ and $(x_2,y_2)$ must necessarily be the diameter of $C$.

Consider an arbitrary $(x_1, y_1) \in C$. Next, let the function $F_{x_1,y_1}$ be defined as $F_{x_1,y_1}: C \to \mathbb R^+_0$, where $F_{x_1,y_1}\left((x,y)\right)=\sqrt{(x-x_1)^2+(y-y_1)^2}\quad (*_1)$.

It will be more convenient to parameterize $x_1,y_1, y, $and $y$. To do this, we will need the two sets $S$ and $T$ defined as:

$$S=\left\{x \in \mathbb R: \exists y \left[(x,y)\in C\right] \right\}$$

$$T=\left\{y \in \mathbb R: \exists x \left[(x,y)\in C\right] \right\}$$

Next, consider the functions $g$ and $h$ defined as:

$$g:\mathbb R \to S, \text{ such that }g(\theta)=\cos(\theta)r_0+x_0$$

$$h:\mathbb R \to T, \text{ such that }h(\theta)=\sin(\theta)r_0+y_0$$

It is straightforward to confirm that for any $\theta \in \mathbb R$, $\left(g(\theta),h(\theta)\right) \in C$.Now, we can parameterize as follows:

\begin{align} &(1)\quad g(\theta_1)=x_1=\cos(\theta_1)r_0+x_0\\ &(2)\quad h(\theta_1)=y_1=\sin(\theta_1)r_0+y_0\\&(3)\quad g(\theta)=x=\cos(\theta)r_0+x_0 \\ &(4)\quad h(\theta)=y=\sin(\theta)r_0+y_0\end{align}

Plugging in these values for $(*_1)$ gives us:

\begin{align}F_{x_1,y_1}\left(\left(g(\theta),h(\theta)\right)\right)&=\sqrt{(\cos(\theta)r_0-\cos(\theta_1)r_0)^2+(\sin(\theta)r_0-\sin(\theta_1)r_0)^2}\\&=\sqrt{r_0^2\left(\cos^2(\theta)+\cos^2(\theta_1)-2\cos(\theta)\cos(\theta_1)\right)+r_0^2\left(\sin^2(\theta)+\sin^2(\theta_1)-2\sin(\theta)\sin(\theta_1)\right)}\\ &=r_0\sqrt{2}\cdot\sqrt{1-\cos(\theta)\cos(\theta_1)-\sin(\theta)\sin(\theta_1)} \end{align}

Next, take the derivative of this expression with respect to $\theta$. Applying the chain rule we have:

\begin{align}&\left[F_{x_1,y_1}\left(\left(g(\cdot),h(\cdot)\right)\right)\right]'(\theta) &=\frac{r_0\sqrt{2}}{\sqrt{1-\cos(\theta)\cos(\theta_1)-\sin(\theta)\sin(\theta_1)}}\cdot\left [\sin(\theta)\cos(\theta_1)-\cos(\theta)\sin(\theta_1)\right]\end{align}

In order to find points where $F_{x_1,y_1}$ attains its maximum, we will set the numerator to $0$:

\begin{align}0&=\frac{r_0\sqrt{2}}{\sqrt{1-\cos(\theta)\cos(\theta_1)-\sin(\theta)\sin(\theta_1)}}\cdot\left [\sin(\theta)\cos(\theta_1)-\cos(\theta)\sin(\theta_1)\right] \\\sin(\theta)\cos(\theta_1)&=\cos(\theta)\sin(\theta_1) \end{align}

At this point, we can break down this expression in to several cases:

  1. $\sin(\theta_1) = 0$
  2. $\cos(\theta_1) = 0$
  3. $\sin(\theta_1) \neq 0$ and $\cos(\theta_1) \neq 0$

Case 1: Suppose $\sin(\theta_1) = 0$. Then $\cos(\theta_1)=0$ or $\sin(\theta)=0$. Clearly, $\cos(\theta_1) \neq 0$ because no such value of $\theta$ has a $\sin$ value of $0$ and $\cos$ value of $0$ simultaneously $\color{red}{(\dagger)}$: so $\sin(\theta)=0$. This means that $\theta = 0\pm 2\pi n$ or $\theta = \pi \pm 2\pi n$.

Case 2: A similar argument will show that we must have $\theta=\frac{\pi}{2} \pm 2\pi n$ or $\theta = \frac{3\pi}{2} \pm 2\pi n$.

Case 3: Suppose $\sin(\theta_1) \neq 0$ and $\cos(\theta_1) \neq 0$. In this case, we write our expression as:

$$\sin(\theta)=\cos(\theta)\cdot \frac{\sin(\theta_1)}{\cos(\theta_1)}$$

We can quickly rule out the subcases of i) $\sin(\theta)=0$ and ii) $\cos(\theta)=0$. Otherwise, we will have a similar contradiction as the point raised for $\color{red}{(\dagger)}$. So we must have $\sin(\theta) \neq 0$ and $\cos(\theta)\neq 0$. Therefore, we can rewrite our equation as:

\begin{align}\frac{\sin(\theta)}{\cos(\theta)}&=\frac{\sin(\theta_1)}{\cos(\theta_1)} \\ \tan(\theta)&=\tan(\theta_1)\end{align}

The $\tan$ function oscillates at a period of $\pi$. Therefore $\theta=\theta_1 \pm \pi n$.

Returning to each case, we will now show where the maximum point is located for a given $\theta_1$...which is related to the arbitrary ordered pair $(x_1,y_1)$ through $g$ and $h$ as $(\cos(\theta_1)r_0+x_0,\sin(\theta_1)r_0+y_0)$. Note that because $\cos$ and $\sin$ have periods of $2\pi$, we will only consider when $n=0$.

Case 1: Comparing $0$ Versus $\pi$ when $\sin(\theta_1)=0$

$\sin(\theta_1)=0$ for two different values: $0$ and $\pi$. Supposing $\theta_1=0$, plugging our values into $F_{x_1,y_1}$ yields:

\begin{align}&(1)\quad F_{x_1,y_1}\left(0\right)=0 \\&(2)\quad F_{x_1,y_1}\left(\pi\right)=2r_0\end{align}

Supposing $\theta_1=\pi$, plugging our values into $F_{x_1,y_1}$ yields:

\begin{align}&(1)\quad F_{x_1,y_1}\left(0\right)=2r_0\\&(2)\quad F_{x_1,y_1}\left(\pi\right)=0\end{align}

Case 2: Comparing $\frac{\pi}{2}$ Versus $\frac{3\pi}{2}$ when $\cos(\theta_1)=0$

A similar argument can be made here, showing the following:

Supposing $\theta_1=\frac{\pi}{2}$, plugging our values into $F_{x_1,y_1}$ yields:

\begin{align}&(1)\quad F_{x_1,y_1}\left(\frac{\pi}{2}\right)=0 \\&(2)\quad F_{x_1,y_1}\left(\frac{3\pi}{2}\right)=2r_0\end{align}

Supposing $\theta_1=\frac{3\pi}{2}$, plugging our values into $F_{x_1,y_1}$ yields:

\begin{align}&(1)\quad F_{x_1,y_1}\left(\frac{\pi}{2}\right)=2r_0 \\&(2)\quad F_{x_1,y_1}\left(\frac{3\pi}{2}\right)=0\end{align}

Case 3: Comparing $\theta_1$ Versus $\theta_1+\pi$ when $\sin(\theta_1) \neq 0$ and $\cos(\theta_1)\neq 0$

Noting that $\cos(\theta+\pi)=-\cos(\theta)$ and $\sin(\theta+\pi)=-\sin(\theta)$, we have the following:

\begin{align}&(1)\quad F_{x_1,y_1}\left(\theta_1\right)=0 \\&(2)\quad F_{x_1,y_1}\left(\theta_1+\pi\right)=2r_0\end{align}

As can be seen, across all possible cases, the ordered pair laying on the circumference of the circle that maximizes the Eulcidean distance to some other arbitrary point $(x_1,y_1)$ on the circumference is precisely the point that is $180 \deg$ ($\pi \text{ rad})$ further along on the circumference. In particular, then, if $(x_1,y_1)=\left(\cos(\theta_1)r_0+x_0,\sin(\theta_1)r_0+y_0 \right)$, then the maximizing point is $\left(x_2,y_2 \right)=\left(\cos(\theta_1+\pi)r_0+x_0,\sin(\theta_1+\pi)r_0+y_0 \right)$

From this post here (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), the line segment connecting $(x_1,y_1)$ to $(x_2,y_2)$ must pass through the center of the circle $(x_0,y_0)$. But this is precisely the definition of the diameter of the circle. Therefore, we see that the diameter is, in fact, the longest chord on the circle $C$.

Importantly, revisiting the expression for the derivative $[F_{x_1,y_1}]'$, we know that the function is not defined when $1-\cos(\theta)\cos(\theta_1)-\sin(\theta)\sin(\theta_1)=0$.

Taken from here (Determine at which values of $\theta$ the function $f(\theta)=\cos(\theta)\cos(\theta_1)+\sin(\theta)\sin(\theta_1)$ equals $1$), this occurs when $\theta=\theta_1$. This means that we cannot rule out that the maximum occurs at $\theta=\theta_1$. However, if we go back up to our previous section, we note that any time $\theta=\theta_1$, we have $F_{x_1,y_1}=0$, which is certainly less than $2$. Given that $F_{x_1,y_1}$ is differentiable every but $\theta_1$, all other possible local maximum's would have been identified by our approach...but the only maximum values we identified were $2r_0$, all of which occurred exclusively when $\theta=\theta_1+\pi$.

We can then conclude that for a given $(x_1,y_1)$ pair, there is uniquely one other pair $(x_2,y_2)$ on the same circle that maximizes the Euclidean distance. And this point is the $180 \deg$ rotated version of $(x_1,y_1)$.

  • 4
    $\begingroup$ The triangle inequality immediately shows that the distance between two points on the circle it at most twice the radius. $\endgroup$
    – Martin R
    Mar 23, 2022 at 9:19
  • 1
    $\begingroup$ WLOG the circle is a unit circle in the complex plane, and one of the ends of the chord goes through $1$, and the other through some $z$ of length 1. The length of the chord is $|1 - z|$, which is at most 2 by the triangle inequality. Taking $z = -1$ actually achieves $|1 - z| = 2$. $\endgroup$
    – Joppy
    Mar 23, 2022 at 9:20

1 Answer 1


$$(x;y)\in C \Leftrightarrow \exists t\in \mathbb{R}: x=x_0+r_0 \cos t \land y=y_0+r_0\sin t$$

Proof can be done with considering two cases. Case 1: $x\geq x_0$, take $t=\arcsin\frac{y-y_0}{r_0}$. Case 2: $x<x_0$, take $t=\pi-\arcsin\frac{y-y_0}{r_0}$.

Let $x_1=x_0+r_0\cos t_1$, $y_1=y_0+r_0\sin t_1$, $x_2=x_0+r_0\cos t_2$, $y_2=y_0+r_0\sin t_2$.

Then distance $$d=r_0 \sqrt{(\cos t_1-\cos t_2)^2+(\sin t_1-\sin t_2)^2}=r_0 \sqrt{2-2\cos (t_1-t_2)}$$

Maximum possible distance corresponds to maximum of $2-2\cos (t_1-t_2)$. Maximum value of $d$ is $d_{max}=2r_0$ and is obtained at $t_1-t_2=\pi+2k\pi$, $k\in\mathbb{Z}$.

At $t_1=t_2+\pi+2k\pi$: $x_1=x_0-r_0 \cos t_2$, $y_1=y_0-r_0 \sin t_2$, $x_0=\frac{x_1+x_2}{2}$, $y_0=\frac{y_1+y_2}{2}$, so center of circle $(x_0;y_0)$ is the middle of chord, that's why maximum length chord is diameter.


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