Is this quadrilateral cyclic?
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.