How to get the equation where a circle goes through three points

The given points are vertices of an isosceles right triangle,

$O(0,0)$ is the center of its hypotenuse $PQ$, and $|PQ|=2\sqrt2$.

Thus is the equation of the circumference $$x^2+y^2=2.$$

A clever strategy (which I got from this generalization) is to simplify the following determinant equation: $$ \det\begin{bmatrix} x^2+y^2 & x & y & 1 \\ x_0^2+y_0^2 & x_0 & y_0 & 1 \\ x_1^2+y_1^2 & x_1 & y_1 & 1 \\ x_2^2+y_2^2 & x_2 & y_2 & 1 \\ \end{bmatrix} = 0 $$ where $(x_0,y_0) = (1,1)$, $(x_1,y_1) = (-1,-1)$, and $(x_2,y_2) = (-1,1)$ are the three points.

This is the equation of a circle because the first row of the matrix is the only one that depends on $x$ and $y$; the determinant is linear in each row, so the equation will have the form $a(x^2+y^2) + bx + cy + d = 0$.

(It's possible that $a = 0$, in which case we'll get a line: a degenerate circle. But this only happens when we are given three collinear points.)

This circle passes through all three points, because when we set $x = x_i$ and $y = y_i$, the matrix has two identical rows, so it's singular and the determinant is $0$.

In this problem, by plotting the three points, it might become easy to find the circle by hand, but this method works well in general.

Dividing by $a\neq 0$ we obtain

$$ax^2+ay^2+bx+cy+d=0\iff x^2+y^2+\frac b a x+\frac c a y+ \frac d a=0$$

then use the three given condition to find $b/a$, $c/a$ and $d/a$.