# Analytical proof that the diameter is the longest chord of a circle

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 $$(x_1,y_1)$$.

• The triangle inequality immediately shows that the distance between two points on the circle it at most twice the radius. Mar 23, 2022 at 9:19
• 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 $|1 - z| = 2$. Mar 23, 2022 at 9:20

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