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.