How to determine whether 4 points are concyclic

Standard

Problem

Given 4 points $a,b,c,d$ in plane, how to determine they fall on a common circle?
There are lots of methods can do this, and here I introduce the determinant method, which is very straightforward, elegant and can easily generalized to quadratic curve.
The following content is inspired by this.
First of all, we can write the circle equation defined by $a,b,c$ by a determinant:
$D:=\begin{vmatrix} a_x & a_y & a_x^2+a_y^2 & 1 \\ b_x & b_y & b_x^2+b_y^2 & 1 \\ c_x & c_y & c_x^2+c_y^2 & 1 \\ p_x & p_y & p_x^2+p_y^2 & 1 \end{vmatrix}=0$

Proposition 1:

When $D>0$, $p$ is inside the circle defined by $a,b,c$，and outside when $D<0$ , right on the circle when $D=0$,which means a circle go through $a,b,c,p$ simultaneously.

Proof:
Notice that:
$D = -p_x M_{41} + p_y M_{42} - (p_x^2 +p_y^2) M_{43} + M_{44}$
($M_{ij}$ is the minor of the matrix)
and if we regard $p$ as a moving point, then $D=0$ is actually an equation of a circle:

$(p_x+\frac{1}{2}\frac{M_{41}}{M_{43}})^2+(p_y-\frac{1}{2}\frac{M_{42}}{M_{43}})^2=(\frac{1}{2}\frac{M_{41}}{M_{43}})^2+(\frac{1}{2}\frac{M_{42}}{M_{43}})^2+\frac{M_{44}}{M_{43}}$

i.e. the circle locate at $(x_0,y_0)=(\frac{1}{2}\frac{M_{41}}{M_{43}},-\frac{1}{2}\frac{M_{42}}{M_{43}})$
with radius $r_0^2=x_0^2+y_0^2+\frac{M_{44}}{M_{43}}$.

When $D>0$ the circle equation become $(x-x_0)^2+(y-y_0)^2

So it’s clear that if $(p_x,p_y)$ fall inside that circle, then $D>0$ and vice versa.
$\square$

Note: According to Cramer’s rule, $(-\frac{M_{41}}{M_{43}},\frac{M_{42}}{M_{43}},\frac{M_{44}}{M_{43}})$ is exactly the solution of the equation
$\begin{pmatrix} a_x & a_y & 1 \\ b_x & b_y & 1 \\ c_x & c_y & 1 \end{pmatrix} \boldsymbol{x}=\begin{pmatrix}a_x^2+a_y^2 \\ b_x^2+b_y^2 \\c_x^2+c_y^2\end{pmatrix}$

And this 3 parameters perfectly define a circle pass through $a,b,c$.One can easily verify when $p$ is $a,b$ or $c$, $D=0$ by the equation above.

Generalization

We can generalize this method to the question of determine whether 6 points $a,b,c,d,e,f$ lies on some quadratic curve $\mathcal{C}$
Any equation of quadratic curve can write as (and again, we regard $f$ as a moving point $p$)
$D:=\begin{vmatrix} a_x^2 & a_x a_y & a_y^2 &a_x & a_y & 1 \\ b_x^2 & b_x b_y & b_y^2 &b_x & b_y & 1 \\ c_x^2 & c_x c_y & c_y^2 &c_x & c_y & 1 \\ d_x^2 & d_x d_y & d_y^2 &d_x & d_y & 1 \\ e_x^2 & e_x e_y & e_y^2 &e_x & e_y & 1 \\ p_x^2 & p_x p_y & p_y^2 &p_x & p_y & 1 \\ \end{vmatrix}=0$
and expand the last row of $D$:
$D=−p_x^2 M_{61}+p_x p_y M_{62}−p_y^2 M_{63}+p_x M_{64}−p_y M_{65}+M_{66}$

Definition.
A point is inside of a quadratic curve $\mathcal{C}$ if there are no tangent of that curve pass through $p$. And outside if there is at least 1 tangent pass through it.

And still we can tell whether $p$ lies in or out of $\mathcal{C}$ by check the sign of $D$.
Let $A,B,C,D,E,F$ be the coefficient of the quadratic equation, we have:

Proposition
The point $p$ lies inside of a quadratic curve $\mathcal{C}$ If $D$ has the same sign with the determinant
$\begin{vmatrix}A & \frac{1}{2}B & \frac{1}{2}D\\ \frac{1}{2}B & C & \frac{1}{2}E \\ \frac{1}{2}D & \frac{1}{2}E & F \end{vmatrix}$

But unfortunately I can’t give the proof temporarily. If you have any idea, please let me know, thanks in ahead!

Check the shape of the curve

If we want to check whether the six point lies on eclipse or hyperbola etc. we can simply check the sign of geometry invariants

Symbols.
$I_1=A+C$
$I_2=\begin{vmatrix} A & \frac{1}{2}B \\\ \frac{1}{2}B & C\end{vmatrix}$
$I_3=\begin{vmatrix}A & \frac{1}{2}B & \frac{1}{2}D\\\ \frac{1}{2}B & C & \frac{1}{2}E \\\ \frac{1}{2}D & \frac{1}{2}E & F \end{vmatrix}$
$K_1=\begin{vmatrix} A & \frac{1}{2}D \\\ \frac{1}{2}D & F\end{vmatrix} + \begin{vmatrix} C & \frac{1}{2}E \\\ \frac{1}{2}E & F\end{vmatrix}$

Theorem.
The curve is classified by the following criterion:
Eclipse : $I_2>0,I_1 I_3<0$
Parabola: $I_2=0,I_3\neq 0$
Hyperbola: $I_2<0,I_3\neq 0$
Two parallel lines : $I_2=I_3,K_1<0$
Two intersecting lines: $I_2<0,I_3=0$
A line( two line coincide): $I_2=I_3=K_1=0$
Imaginary ellipse :$I_2>0, I_1 I_3 >0$
Point(two im. line intersect at real plane) : $I_2>0,I_3=0$
Two parallel imaginary lines: $I_2=I_3=0,K_1>0$