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 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), 292825 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.

### Port of Misha Lavrov's TI-Basic solution, 33 bytes

{![*](map */*,(\$_ Z-.rotate)).im}


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• 42? Is it still accurate? – Jo King Nov 18 '18 at 4:24
• @JoKing No, it's not. – nwellnhof Nov 18 '18 at 9:44
• What does the colon do in this case? It's definitely not a label, and also not a method call. – user202729 Nov 18 '18 at 14:55
• @user202729 It is a method call with indirect invocant syntax. – nwellnhof Nov 18 '18 at 15:44

# 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.

### JavaScript (ES6) (101 bytes)

p=>(h=(a,b)=>Math.hypot(p[a]-p[b],p[a+1]-p[b+1]))&&((h(2,4)*h(0,6)+h(0,2)*h(4,6)-h(0,4)*h(2,6))<1e-8)


Takes input as [x1,y1,x2,y2,x3,y3,x4,y4], outputs a Boolean.

Checked based on $$ef=ac+bd$$ where $$\e,f\$$ are the diagonals and $$\a,b,c,d\$$ are the sides in order.

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# Jelly, 11 bytes

²Sṭ;L€€ṖÆḊ¬


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Uses the determinant approach from Misha Lavrov's Mathematica solution. Outputs 1 for true, 0 for false.

### How it works

²Sṭ;L€€ṖÆḊ¬  Main link (monad). Input: [[x1,x2,x3,x4], [y1,y2,y3,y4]]
²S           Square each scalar and add row-wise; [x1*x1+y1*y1, ...]
ṭ          Append to the input
;L€€      Add two rows of [1,1,1,1]'s
Ṗ     Remove an extra row
ÆḊ¬  Is the determinant zero?


# Jelly, 12 bytes

Iµ÷×ƭ/÷SµḞ=A


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Uses the convoluted cross-ratio approach from Misha Lavrov's TI-Basic solution. Outputs 1 for true, 0 for false.

### How it works

Iµ÷×ƭ/÷SµḞ=A  Main link (monad). Input: list of four complex numbers [z1,z2,z3,z4]
I             Increments; [z2-z1, z3-z2, z4-z3]
µ            Refocus on above for sum function
÷×ƭ/÷S      (z2-z1)÷(z3-z2)×(z4-z3)÷(z4-z1)
µ     Refocus again
Ḟ=A  (real part) == (norm) within error margin
i.e. imag part is negligible?


I believe both are golfable...

# APL (Dyalog Classic), 25 bytes

{0=-/|⍵}(-⌿2 3⍴2/⌽)×⊃-1↓⊢


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Ptolemy's theorem, credit: Кирилл Малышев's answer