Circumcircle

In the Euclidean plane, it is possible to give explicitly an equation of the circumcircle in terms of the Cartesian coordinates of the vertices of the inscribed triangle. Suppose that

${\displaystyle {\begin{aligned}\mathbf {A} &=(A_{x},A_{y})\\\mathbf {B} &=(B_{x},B_{y})\\\mathbf {C} &=(C_{x},C_{y})\end{aligned}}}$

are the coordinates of points A, B, C. The circumcircle is then the locus of points ${\displaystyle \mathbf {v} =(v_{x},v_{y})}$ in the Cartesian plane satisfying the equations

${\displaystyle {\begin{aligned}|\mathbf {v} -\mathbf {u} |^{2}&=r^{2}\\|\mathbf {A} -\mathbf {u} |^{2}&=r^{2}\\|\mathbf {B} -\mathbf {u} |^{2}&=r^{2}\\|\mathbf {C} -\mathbf {u} |^{2}&=r^{2}\end{aligned}}}$

guaranteeing that the points A, B, C, v are all the same distance r from the common center ${\displaystyle \mathbf {u} }$ of the circle. Using the polarization identity, these equations reduce to the condition that the matrix

${\displaystyle {\begin{bmatrix}|\mathbf {v} |^{2}&-2v_{x}&-2v_{y}&-1\\|\mathbf {A} |^{2}&-2A_{x}&-2A_{y}&-1\\|\mathbf {B} |^{2}&-2B_{x}&-2B_{y}&-1\\|\mathbf {C} |^{2}&-2C_{x}&-2C_{y}&-1\end{bmatrix}}}$

has a nonzero kernel. Thus the circumcircle may alternatively be described as the locus of zeros of the determinant of this matrix:

${\displaystyle \det {\begin{bmatrix}|\mathbf {v} |^{2}&v_{x}&v_{y}&1\\|\mathbf {A} |^{2}&A_{x}&A_{y}&1\\|\mathbf {B} |^{2}&B_{x}&B_{y}&1\\|\mathbf {C} |^{2}&C_{x}&C_{y}&1\end{bmatrix}}=0.}$

Using cofactor expansion, let

${\displaystyle {\begin{aligned}S_{x}&={\frac {1}{2}}\det {\begin{bmatrix}|\mathbf {A} |^{2}&A_{y}&1\\|\mathbf {B} |^{2}&B_{y}&1\\|\mathbf {C} |^{2}&C_{y}&1\end{bmatrix}},\\[5pt]S_{y}&={\frac {1}{2}}\det {\begin{bmatrix}A_{x}&|\mathbf {A} |^{2}&1\\B_{x}&|\mathbf {B} |^{2}&1\\C_{x}&|\mathbf {C} |^{2}&1\end{bmatrix}},\\[5pt]a&=\det {\begin{bmatrix}A_{x}&A_{y}&1\\B_{x}&B_{y}&1\\C_{x}&C_{y}&1\end{bmatrix}},\\[5pt]b&=\det {\begin{bmatrix}A_{x}&A_{y}&|\mathbf {A} |^{2}\\B_{x}&B_{y}&|\mathbf {B} |^{2}\\C_{x}&C_{y}&|\mathbf {C} |^{2}\end{bmatrix}}\end{aligned}}}$

we then have ${\displaystyle a|\mathbf {v} |^{2}-2\mathbf {Sv} -b=0}$ where ${\displaystyle \mathbf {S} =(S_{x},S_{y}),}$ and – assuming the three points were not in a line (otherwise the circumcircle is that line that can also be seen as a generalized circle with S at infinity) – ${\displaystyle \left|\mathbf {v} -{\tfrac {\mathbf {S} }{a}}\right|^{2}={\tfrac {b}{a}}+{\tfrac {|\mathbf {S} |^{2}}{a^{2}}},}$ giving the circumcenter ${\displaystyle {\tfrac {\mathbf {S} }{a}}}$ and the circumradius ${\displaystyle {\sqrt {{\tfrac {b}{a}}+{\tfrac {|\mathbf {S} |^{2}}{a^{2}}}}}.}$ A similar approach allows one to deduce the equation of the circumsphere of a tetrahedron.

A unit vector perpendicular to the plane containing the circle is given by

${\displaystyle {\widehat {n}}={\frac {(P_{2}-P_{1})\times (P_{3}-P_{1})}{|(P_{2}-P_{1})\times (P_{3}-P_{1})|}}.}$

Hence, given the radius, r, center, P_c, a point on the circle, P₀ and a unit normal of the plane containing the circle, ${\displaystyle {\widehat {n}},}$ one parametric equation of the circle starting from the point P₀ and proceeding in a positively oriented (i.e., right-handed) sense about ${\displaystyle {\widehat {n}}}$ is the following:

${\displaystyle \mathrm {R} (s)=\mathrm {P_{c}} +\cos \left({\frac {\mathrm {s} }{\mathrm {r} }}\right)(P_{0}-P_{c})+\sin \left({\frac {\mathrm {s} }{\mathrm {r} }}\right)\left[{\widehat {n}}\times (P_{0}-P_{c})\right].}$

An equation for the circumcircle in trilinear coordinates x : y : z is[1] ${\displaystyle {\tfrac {a}{x}}+{\tfrac {b}{y}}+{\tfrac {c}{z}}=0.}$ An equation for the circumcircle in barycentric coordinates x : y : z is ${\displaystyle {\tfrac {a^{2}}{x}}+{\tfrac {b^{2}}{y}}+{\tfrac {c^{2}}{z}}=0.}$

The isogonal conjugate of the circumcircle is the line at infinity, given in trilinear coordinates by ${\displaystyle ax+by+cz=0}$ and in barycentric coordinates by ${\displaystyle x+y+z=0.}$

Additionally, the circumcircle of a triangle embedded in d dimensions can be found using a generalized method. Let A, B, C be d-dimensional points, which form the vertices of a triangle. We start by transposing the system to place C at the origin:

${\displaystyle {\begin{aligned}\mathbf {a} &=\mathbf {A} -\mathbf {C} ,\\\mathbf {b} &=\mathbf {B} -\mathbf {C} .\end{aligned}}}$

The circumradius r is then

${\displaystyle r={\frac {\left\|\mathbf {a} \right\|\left\|\mathbf {b} \right\|\left\|\mathbf {a} -\mathbf {b} \right\|}{2\left\|\mathbf {a} \times \mathbf {b} \right\|}}={\frac {\left\|\mathbf {a} -\mathbf {b} \right\|}{2\sin \theta }}={\frac {\left\|\mathbf {A} -\mathbf {B} \right\|}{2\sin \theta }},}$

where θ is the interior angle between a and b. The circumcenter, p₀, is given by

${\displaystyle p_{0}={\frac {(\left\|\mathbf {a} \right\|^{2}\mathbf {b} -\left\|\mathbf {b} \right\|^{2}\mathbf {a} )\times (\mathbf {a} \times \mathbf {b} )}{2\left\|\mathbf {a} \times \mathbf {b} \right\|^{2}}}+\mathbf {C} .}$

This formula only works in three dimensions as the cross product is not defined in other dimensions, but it can be generalized to the other dimensions by replacing the cross products with following identities:

${\displaystyle {\begin{aligned}(\mathbf {u} \times \mathbf {v} )\times \mathbf {w} &=(\mathbf {u} \cdot \mathbf {w} )\mathbf {v} -(\mathbf {v} \cdot \mathbf {w} )\mathbf {u} ,\\\mathbf {u} \times (\mathbf {v} \times \mathbf {w} )&=(\mathbf {u} \cdot \mathbf {w} )\mathbf {v} -(\mathbf {u} \cdot \mathbf {v} )\mathbf {w} ,\\\left\|\mathbf {u} \times \mathbf {v} \right\|^{2}&=\left\|\mathbf {u} \right\|^{2}\left\|\mathbf {v} \right\|^{2}-(\mathbf {u} \cdot \mathbf {v} )^{2}.\end{aligned}}}$

The Cartesian coordinates of the circumcenter ${\displaystyle U=\left(U_{x},U_{y}\right)}$ are

${\displaystyle {\begin{aligned}U_{x}&={\frac {1}{D}}\left[(A_{x}^{2}+A_{y}^{2})(B_{y}-C_{y})+(B_{x}^{2}+B_{y}^{2})(C_{y}-A_{y})+(C_{x}^{2}+C_{y}^{2})(A_{y}-B_{y})\right]\\[5pt]U_{y}&={\frac {1}{D}}\left[(A_{x}^{2}+A_{y}^{2})(C_{x}-B_{x})+(B_{x}^{2}+B_{y}^{2})(A_{x}-C_{x})+(C_{x}^{2}+C_{y}^{2})(B_{x}-A_{x})\right]\end{aligned}}}$

with

${\displaystyle D=2\left[A_{x}(B_{y}-C_{y})+B_{x}(C_{y}-A_{y})+C_{x}(A_{y}-B_{y})\right].\,}$

Without loss of generality this can be expressed in a simplified form after translation of the vertex A to the origin of the Cartesian coordinate systems, i.e., when ${\displaystyle A'=A-A=(A'_{x},A'_{y})=(0,0).}$ In this case, the coordinates of the vertices ${\displaystyle B'=B-A}$ and ${\displaystyle C'=C-A}$ represent the vectors from vertex A' to these vertices. Observe that this trivial translation is possible for all triangles and the circumcenter ${\displaystyle U'=(U'_{x},U'_{y})}$ of the triangle △A'B'C' follow as

${\displaystyle {\begin{aligned}U'_{x}&={\frac {1}{D'}}\left[C'_{y}({B'_{x}}^{2}+{B'_{y}}^{2})-B'_{y}({C'_{x}}^{2}+{C'_{y}}^{2})\right],\\[5pt]U'_{y}&={\frac {1}{D'}}\left[B'_{x}({C'_{x}}^{2}+{C'_{y}}^{2})-C'_{x}({B'_{x}}^{2}+{B'_{y}}^{2})\right]\end{aligned}}}$

with

${\displaystyle D'=2(B'_{x}C'_{y}-B'_{y}C'_{x}).\,}$

Due to the translation of vertex A to the origin, the circumradius r can be computed as

${\displaystyle r=\|U'\|={\sqrt {{U'_{x}}^{2}+{U'_{y}}^{2}}}}$

and the actual circumcenter of △ABC follows as

${\displaystyle U=U'+A}$

The circumcenter has trilinear coordinates[2]

${\displaystyle \cos \alpha$ :\cos \beta :\cos \gamma }

where α, β, γ are the angles of the triangle.

In terms of the side lengths a, b, c, the trilinears are[3]

${\displaystyle a\left(b^{2}+c^{2}-a^{2}\right):b\left(c^{2}+a^{2}-b^{2}\right):c\left(a^{2}+b^{2}-c^{2}\right).}$

The circumcenter has barycentric coordinates[4]

${\displaystyle a^{2}\left(b^{2}+c^{2}-a^{2}\right):\;b^{2}\left(c^{2}+a^{2}-b^{2}\right):\;c^{2}\left(a^{2}+b^{2}-c^{2}\right),\,}$

where a, b, c are edge lengths BC, CA, AB respectively) of the triangle.

In terms of the triangle's angles α, β, γ, the barycentric coordinates of the circumcenter are[3]

${\displaystyle \sin 2\alpha$ :\sin 2\beta :\sin 2\gamma .}

Since the Cartesian coordinates of any point are a weighted average of those of the vertices, with the weights being the point's barycentric coordinates normalized to sum to unity, the circumcenter vector can be written as

${\displaystyle U={\frac {a^{2}\left(b^{2}+c^{2}-a^{2}\right)A+b^{2}\left(c^{2}+a^{2}-b^{2}\right)B+c^{2}\left(a^{2}+b^{2}-c^{2}\right)C}{a^{2}\left(b^{2}+c^{2}-a^{2}\right)+b^{2}\left(c^{2}+a^{2}-b^{2}\right)+c^{2}\left(a^{2}+b^{2}-c^{2}\right)}}.}$

Here U is the vector of the circumcenter and A, B, C are the vertex vectors. The divisor here equals 16S ² where S is the area of the triangle. As stated previously

${\displaystyle {\begin{aligned}\mathbf {a} &=\mathbf {A} -\mathbf {C} ,\\\mathbf {b} &=\mathbf {B} -\mathbf {C} .\end{aligned}}}$

In Euclidean space, there is a unique circle passing through any given three non-collinear points P₁, P₂, P₃. Using Cartesian coordinates to represent these points as spatial vectors, it is possible to use the dot product and cross product to calculate the radius and center of the circle. Let

${\displaystyle \mathrm {P_{1}} ={\begin{bmatrix}x_{1}\\y_{1}\\z_{1}\end{bmatrix}},\mathrm {P_{2}} ={\begin{bmatrix}x_{2}\\y_{2}\\z_{2}\end{bmatrix}},\mathrm {P_{3}} ={\begin{bmatrix}x_{3}\\y_{3}\\z_{3}\end{bmatrix}}}$

Then the radius of the circle is given by

${\displaystyle \mathrm {r} ={\frac {\left|P_{1}-P_{2}\right|\left|P_{2}-P_{3}\right|\left|P_{3}-P_{1}\right|}{2\left|\left(P_{1}-P_{2}\right)\times \left(P_{2}-P_{3}\right)\right|}}}$

The center of the circle is given by the linear combination

${\displaystyle \mathrm {P_{c}} =\alpha \,P_{1}+\beta \,P_{2}+\gamma \,P_{3}}$

where

${\displaystyle {\begin{aligned}\alpha ={\frac {\left|P_{2}-P_{3}\right|^{2}\left(P_{1}-P_{2}\right)\cdot \left(P_{1}-P_{3}\right)}{2\left|\left(P_{1}-P_{2}\right)\times \left(P_{2}-P_{3}\right)\right|^{2}}}\\\beta ={\frac {\left|P_{1}-P_{3}\right|^{2}\left(P_{2}-P_{1}\right)\cdot \left(P_{2}-P_{3}\right)}{2\left|\left(P_{1}-P_{2}\right)\times \left(P_{2}-P_{3}\right)\right|^{2}}}\\\gamma ={\frac {\left|P_{1}-P_{2}\right|^{2}\left(P_{3}-P_{1}\right)\cdot \left(P_{3}-P_{2}\right)}{2\left|\left(P_{1}-P_{2}\right)\times \left(P_{2}-P_{3}\right)\right|^{2}}}\end{aligned}}}$

The circumcenter's position depends on the type of triangle:

• For an acute triangle (all angles smaller than a right angle), the circumcenter always lies inside the triangle.
• For a right triangle, the circumcenter always lies at the midpoint of the hypotenuse. This is one form of Thales' theorem.
• For an obtuse triangle (a triangle with one angle bigger than a right angle), the circumcenter always lies outside the triangle.

The circumcenter of an acute triangle is inside the triangle

The circumcenter of a right triangle is at the midpoint of the hypotenuse

The circumcenter of an obtuse triangle is outside the triangle

These locational features can be seen by considering the trilinear or barycentric coordinates given above for the circumcenter: all three coordinates are positive for any interior point, at least one coordinate is negative for any exterior point, and one coordinate is zero and two are positive for a non-vertex point on a side of the triangle.

• Weisstein, Eric W. "Circumcircle". MathWorld.
• Weisstein, Eric W. "Cyclic Polygon". MathWorld.
• Weisstein, Eric W. "Steiner circumellipse". MathWorld.

• Triangle circumcircle and circumcenter With interactive animation
• An interactive Java applet for the circumcenter