Simplifying a parametric equation for the formula of a circle given to point on opposite sides

Question

Given two points that are 180 degrees from each other at points $(x_1,y_1)$ and $(x_2,y_2)$ respectively you can find a circle with all points of the form $$x=\cos \left(t\right)^2\left(\tan \left(t\right)\left(x_2\tan \left(t\right)-y_2+y_1\right)+x_1\right)$$ $$y=\sin \left(t\right)^2\left(x_2\tan \left(t\right)-y_2+y_1\right)+\frac{1}{2}\sin \left(2t\right)x_1-x_2\tan \left(t\right)+y_2$$ $$0\le t\le \pi $$

Obviously this is kinda a monster of a parametric set of equations and I was wondering if anyone had any advice on how to go along solving simplifying it, ideally in in a $y=x$ form or something similar.

Answer 1

If these two points are $180º$ from each other, then they are on opposite sides of the circle.

Therefore, the circle's center is $\displaystyle \left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)$.

The radius of the circle would then be $\displaystyle \frac{1}{2}\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

You would then have

$\displaystyle x(t)=\frac{1}{2}(x_1+x_2)+\frac{1}{2}\cos(t)\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

$\displaystyle y(t)=\frac{1}{2}(y_1+y_2)+\frac{1}{2}\sin(t)\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

Update: In order to fit the constraints on $t$, changing the argument of $\sin$ and $\cos$ from $t$ to $2t$ works

$\boxed{\displaystyle x(t)=\frac{1}{2}(x_1+x_2)+\frac{1}{2}\cos(2t)\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}}$

$\boxed{\displaystyle y(t)=\frac{1}{2}(y_1+y_2)+\frac{1}{2}\sin(2t)\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}}$

Answer 2

\begin{eqnarray} x&=&\cos^2 \left(t\right)\left(\tan \left(t\right)\left(x_2\tan \left(t\right)-y_2+y_1\right)+x_1\right)\\ y&=&\sin^2 \left(t\right)\left(x_2\tan \left(t\right)-y_2+y_1\right)+\frac{1}{2}x_1\sin \left(2t\right)-x_2\tan \left(t\right)+y_2 \end{eqnarray}

The points $(x_1,y_1)$ and $(x_2,y_2)$ lie at opposite ends of the diameter of a circle of radius $r$ and center $(h,k)=\left(\dfrac{x_1+x_2}{2},\dfrac{y_1+y_2}{2}\right)$.

Thus the translation $\left(\dfrac{x_1+x_2}{2},\dfrac{y_1+y_2}{2}\right)\rightarrow(0,0)$ moves the circle to a circle with radius $r$ about the origin and

\begin{eqnarray} (x_1,y_1)\rightarrow\left(\frac{x_1-x_2}{2},\frac{y_1-y_2}{2}\right)=(u,v)\\ (x_2,y_2)\rightarrow\left(-\frac{x_1-x_2}{2},-\frac{y_1-y_2}{2}\right)=(-u,-v)\\ \end{eqnarray}

This translates equations $(1)$ and $(2)$ as follows

\begin{eqnarray} x&=&\phantom{-1}u\cos(2t)+v\sin(2t)\\ y&=&-v\cos(2t)+u\sin(2t) \end{eqnarray}

Translating back to the original position gives

\begin{eqnarray} x&=&\phantom{-1}u\cos(2t)+v\sin(2t)+h\\ y&=&-v\cos(2t)+u\sin(2t)+k \end{eqnarray}

Replacing the original variables gives

\begin{eqnarray} x(t)&=&\frac{1}{2}[(x_1+x_2)+(x_1-x_2)\cos(2t)+(y_1-y_2)\sin(2t)]\\ y(t)&=&\frac{1}{2}[(y_1+y_2)-(y_1-y_2)\cos(2t)+(x_1-x_2)\sin(2t)] \end{eqnarray}

Notice that for any $t\in[0,\pi]$ this formulation gives the same results as the original formulation.