Is this quadrilateral cyclic? In mathematics, a cyclic quadrilateral is one whose vertices all lie on the same circle. In other words, every vertex is on the circumcircle of the other three. For more information, see the MathWorld article.
Examples
These quadrilaterals are cyclic:

This trapezoid is not cyclic.

(Images from Wikipedia)
Objective
Given the coordinates of four vertices in counterclockwise order which form a convex quadrilateral, determine if the quadrilateral is cyclic.
Coordinates will be integers (note, however, that the circumcenter coordinates and circumradius are not necessarily integers.) As implied by the previous paragraph, no three points will be co-linear and no two coincident.
I/O
You may take input using any reasonable format. In particular, [[x1,x2,x3,x4],[y1,y2,y3,y4]], [[x1,y1],[x2,y2],[x3,y3],[x4,y4]] and complex numbers are all fine.
Output using any different consistent values for true and false.
Test cases
True:
[0,0], [314,0], [314,1], [0,1]
[-5,5], [5,-5], [1337,42], [42,1337]
[104, -233], [109, -232], [112, -231], [123, -224]

False:
[0,0], [314,0], [314,100], [0,99]
[31,41],[59,26],[53,58],[0,314]
Wolfram Language (Mathematica), 23 bytes

#∈Circumsphere@{##2}&amp;


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Takes four inputs: the lists {x1,y1}, {x2,y2}, {x3,y3}, and {x4,y4}. Checks if the first point lies on the circumcircle of the other three. Also works for checking if \$n+1\$ points in \$\mathbb R^n\$ are concyclic, provided the last \$n\$ of them are affinely independent (because Circumsphere is sad if you give it a degenerate input).

Alternatively, here is a mathematical approach:

Wolfram Language (Mathematica), 29 28 25 24 bytes

Det@{#^2+#2^2,##,1^#}^0&amp;


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Takes two lists as input: {x1,x2,x3,x4} and {y1,y2,y3,y4}. Returns Indeterminate when the four points are on a common circle, and 1 otherwise.

From the four points \$(x_1, y_1), (x_2,y_2), (x_3, y_3), (x_4, y_4)\$, this solution constructs the matrix below:

\$\begin{bmatrix}x_1^2 + y_1^2 &amp; x_2^2 + y_2^2 &amp; x_3^2 + y_3^2 &amp; x_4^2 + y_4^2 \\ x_1 &amp; x_2 &amp; x_3 &amp; x_4 \\ y_1 &amp; y_2 &amp; y_3 &amp; y_4 \\ 1 &amp; 1 &amp; 1 &amp; 1 \end{bmatrix}\$

The determinant of this matrix is 0 if and only if the four rows are linearly dependent, and a linear dependency between the rows is the same thing as the equation of a circle that's satisfied at all four points.

The shortest way I could think of to check if the determinant is 0 is to raise it to the 0-th power: 0^0 is Indeterminate while anything else gives 1.