Three points determine a circle

We have just mentioned how three distinct points determine a circle. It is worth discussing this in a bit more detail. Classical geometry forms a triangle out of three noncollinear points. Then we can talk about the ``circumcircle'' that passes through each of the vertices of the triangle. The center of the circumcircle (called the ``circumcenter'') can be located by finding the intersection of the three perpendicular bisectors of the sides. We will explain this in terms of complex algebra.

Let the three points be $z_1$, $z_2$, $z_3$. Then the circle of center $c$ and radius $r$ that passes through them would satisfy
$$\begin{align*} \vert z_1-c\vert &= \vert z_2-c\vert = \vert z_3-c\vert=r \end{align*}$$
Square these and use complex conjugation to obtain the following equations
$$\begin{align*} z_1 \bar{z_1} - c \bar{z_1} -\bar{c} z_1 +\vert c\vert^2 &= r^2\... ... z_3 \bar{z_3} - c \bar{z_3} -\bar{c} z_3 +\vert c\vert^2 &= r^2\\ \end{align*}$$
Subtract the first equation from the latter two to get two equations
$$\begin{align*} z_2 \bar{z_2}-z_1\bar{z_1} &= c (\bar{z_2}-\bar{z_1}) +\bar{c} (... ...z_3}-z_1\bar{z_1} &= c (\bar{z_3}-\bar{z_1}) +\bar{c} (z_3-z_1) \\ \end{align*}$$
These are the equations of the lines that perpendicularly bisect the sides connecting $z_2$ to $z_1$ and $z_3$ to $z_1$, respectively. These can then be written in matrix form as $$\begin{bmatrix}z_2 \bar{z_2}-z_1\bar{z_1}\\ z_3 \bar{z_3}-... ... & z_3-z_1 \end{bmatrix}\begin{bmatrix}c\\ \bar{c}\end{bmatrix}$$ This has a unique solution if and only if the determinant of the $2\times2$ matrix on the right has nonzero determinant. The determinant is 0 if and only if $$\frac{\bar{z_3}-\bar{z_1}}{z_3-z_1} = \frac{\bar{z_2}-\bar{z_1}}{z_2-z_1} .$$ That is precisely the condition that the three points $z_1$, $z_2$, $z_3$ lie on a line.

Out of these equations we get a formula for the center $c$ of the circle. Plugging this back into any of the original equations then gives a formula for the radius $r$. For the record, here is the formula for the center: $$c = \frac {\vert z_1\vert^2(z_2-z_3) +\vert z_2\vert^2(z_3-z... ...r{z_2}) +z_2(\bar{z_1}-\bar{z_3})+ z_3(\bar{z_2}-\bar{z_1}) }.$$ The radius can be obtained as $\vert z_1-c\vert$.

Note that the denominator is pure imaginary. If it's zero, the three points are collinear. In fact, the denominator may alternatively be written as
$$\begin{align*} d &= {z_1(\bar{z_3}-\bar{z_2}) +z_2(\bar{z_1}-\bar{z_3})+ z_3(\... ...r{(z_2-z_1)}\\ &= 4 i \times\text{Area of triangle $(z_1,z_2,z_3)$} \end{align*}$$
The area has to be interpreted as a signed quantity depending on the orientation of $z_1$, $z_2$, $z_3$.

Although this leads to a less symmetric formula, by using a linear transformation $z\mapsto az+b$ we can simplify the problem of finding the circle passing through three points. First, the transform $$z\to \frac{z-z_1}{z_2-z_1}$$ transforms the points $z_1$, $z_2$, $z_3$ to 0, 1 and $\zeta = \frac{z_3-z_1}{z_2-z_1}$. Then we just have to find the center of the circle passing through 0, 1 and $\zeta$, and then transform back to find the center of the circle through $z_1$, $z_2$, $z_3$. The same algebra as before is a little easier and leads to the center of the circle through 0, 1 and $\zeta$ being $$c'= \frac{\zeta-\vert\zeta\vert^2}{\zeta-\bar{\zeta}}.$$ Then the center through $z_1$, $z_2$ and $z_3$ is
$$\begin{align*} c&= (z_2-z_1)c'+z_1\\ &=(z_2-z_2)\left(\frac{\zeta-\vert\zeta\vert^2}{\zeta-\bar{\zeta}}\right) + z_1\\ \end{align*}$$
Plugging in the formula for $\zeta = \frac{z_3-z_1}{z_2-z_1}$ eventually yields the same formula as before.