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}&

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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&

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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:

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

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.

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.

Python 3, 70 bytes

lambda b,c,d,e,a=abs:a(a(b-d)*a(c-e)-a(b-c)*a(d-e)-a(c-d)*a(b-e))<1e-8

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I use the Ptolemy's theorem.

In a quadrilateral, if the sum of the products of its two pairs of
opposite sides is equal to the product of its diagonals, then the
quadrilateral can be inscribed in a circle.

b, c, d, e are complex numbers.

Perl 6, 44 bytes

{!im ($^b-$^a)*($^d-$^c)/(($d-$a)*($b-$c)):}

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Takes vertices as complex numbers. Uses the fact that the sum of opposite angles is 180° in a cyclic quadrilateral. The order of operations should guarantee that floating-point operations yield an exact result for (small enough) integers.

JavaScript (ES6)

Testing the angles, 114 bytes

Takes input as the array $[x1,y1,x2,y2,x3,y3,x4,y4]$. Returns a Boolean value.

a=>(F=i=>(A=Math.atan2)(a[i+3&7]-(y=a[i+1]),a[i+2&7]-a[i])-A(a[i+5&7]-y,a[i+4&7]-a[i]))(0)+F(2)+F(4)+F(6)==Math.PI

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Computing a determinant, 130 bytes

Takes input as $[x1,x2,x3,x4]$ and $[y1,y2,y3,y4]$ in currying syntax. Returns a Boolean value.

This one is equivalent to MishaLavrov's 2nd answer, with a rotated matrix.

x=>y=>!(g=a=>a+a?a.reduce((v,[r],i)=>v+(i&1?-r:r)*g(a.map(r=>r.slice(1)).filter(_=>i--)),0):1)(x.map((X,i)=>[1,Y=y[i],X,X*X+Y*Y]))

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TI-Basic (83 series), 21 bytes

e^(ΔList(ln(ΔList(augment(Ans,Ans

not(imag(Ans(1)Ans(3

Takes input as a list of four complex numbers in Ans. Returns 1 if the quadrilateral is cyclic and 0 otherwise.

This is nwellnhof's cross-ratio computation, in heavy disguise. If we start with values $z_1, z_2, z_3, z_4$, then:

• 
  ΔList(augment(Ans,Ans computes the differences $z_2-z_1, z_3-z_2, z_4-z_3, z_1-z_4$ (and a few more redundant terms),


• 
  e^(ΔList(ln( of that computes the ratios $frac{z_3-z_2}{z_2-z_1}, frac{z_4-z_3}{z_3-z_2}, frac{z_1-z_4}{z_4-z_3}, dots$.


• We check if the product of the first and third terms, which is $frac{z_3-z_2}{z_2-z_1} cdot frac{z_1-z_4}{z_4-z_3}$, has no imaginary part. Note that this is the same as the cross-ratio $(z_3,z_1;z_2,z_4) = frac{z_2-z_3}{z_2-z_1} : frac{z_4-z_3}{z_4-z_1}$.

I did my best to check if numerical error is a problem, and it doesn't seem to be, but if anyone has good test cases for that, please let me know.