Wolfram Language (Mathematica), 23 bytes
#∈Circumsphere@{##2}&
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), (削除) 29 (削除ここまで) (削除) 28 (削除ここまで) 25 bytes
0==Det@{#^2+#2^2,##,1^#}&
Takes two lists as input: {x1,x2,x3,x4} and {y1,y2,y3,y4}.
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 & x_2^2 + y_2^2 & x_3^2 + y_3^2 & x_4^2 + y_4^2 \\ x_1 & x_2 & x_3 & x_4 \\ y_1 & y_2 & y_3 & y_4 \\ 1 & 1 & 1 & 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.
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