Circumcircle

In geometry, the circumscribed circle or circumcircle of a triangle is a circle that passes through all three vertices. The center of this circle is called the circumcenter of the triangle, and its radius is called the circumradius. The circumcenter is the point of intersection between the three perpendicular bisectors of the triangle's sides, and is a triangle center.

More generally, an Template:Mvar-sided polygon with all its vertices on the same circle, also called the circumscribed circle, is called a cyclic polygon, or in the special case Template:Math, a cyclic quadrilateral. All rectangles, isosceles trapezoids, right kites, and regular polygons are cyclic, but not every polygon is.

Straightedge and compass construction

Construction of the circumcircle of triangle Template:Math and the circumcenter Template:Mvar

The circumcenter of a triangle can be constructed by drawing any two of the three perpendicular bisectors. For three non-collinear points, these two lines cannot be parallel, and the circumcenter is the point where they cross. Any point on the bisector is equidistant from the two points that it bisects, from which it follows that this point, on both bisectors, is equidistant from all three triangle vertices. The circumradius is the distance from it to any of the three vertices. Template:Clear

Alternative construction

Alternative construction of the circumcenter (intersection of broken lines)

An alternative method to determine the circumcenter is to draw any two lines each one departing from one of the vertices at an angle with the common side, the common angle of departure being 90° minus the angle of the opposite vertex. (In the case of the opposite angle being obtuse, drawing a line at a negative angle means going outside the triangle.)

In coastal navigation, a triangle's circumcircle is sometimes used as a way of obtaining a position line using a sextant when no compass is available. The horizontal angle between two landmarks defines the circumcircle upon which the observer lies.

Circumcircle equations

Cartesian coordinates

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

$\begin{array}{cc}𝐀& =(A_x,A_y)\\ 𝐁& =(B_x,B_y)\\ 𝐂& =(C_x,C_y)\end{array}$

are the coordinates of points Template:Mvar. The circumcircle is then the locus of points $𝐯=(v_x,v_y)$ in the Cartesian plane satisfying the equations

$\begin{array}{cc}|𝐯-𝐮{|}^2& =r^2\\ |𝐀-𝐮{|}^2& =r^2\\ |𝐁-𝐮{|}^2& =r^2\\ |𝐂-𝐮{|}^2& =r^2\end{array}$

guaranteeing that the points Template:Math are all the same distance Template:Mvar from the common center $𝐮$ of the circle. Using the polarization identity, these equations reduce to the condition that the matrix

$\left[\begin{array}{cccc}|𝐯{|}^2& -2v_x& -2v_y& -1\\ |𝐀{|}^2& -2A_x& -2A_y& -1\\ |𝐁{|}^2& -2B_x& -2B_y& -1\\ |𝐂{|}^2& -2C_x& -2C_y& -1\end{array}\right]$

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

$\det \left[\begin{array}{cccc}|𝐯{|}^2& v_x& v_y& 1\\ |𝐀{|}^2& A_x& A_y& 1\\ |𝐁{|}^2& B_x& B_y& 1\\ |𝐂{|}^2& C_x& C_y& 1\end{array}\right]=0.$

Using cofactor expansion, let

$\begin{array}{cc}{S}_x& =\frac{1}{2}\det \left[\begin{array}{ccc}|𝐀{|}^2& A_y& 1\\ |𝐁{|}^2& B_y& 1\\ |𝐂{|}^2& C_y& 1\end{array}\right],\\ S_y& =\frac{1}{2}\det \left[\begin{array}{ccc}{A}_x& |𝐀{|}^2& 1\\ B_x& |𝐁{|}^2& 1\\ C_x& |𝐂{|}^2& 1\end{array}\right],\\ a& =\det \left[\begin{array}{ccc}{A}_x& A_y& 1\\ B_x& B_y& 1\\ C_x& C_y& 1\end{array}\right],\\ b& =\det \left[\begin{array}{ccc}{A}_x& A_y& |𝐀{|}^2\\ B_x& B_y& |𝐁{|}^2\\ C_x& C_y& |𝐂{|}^2\end{array}\right]\end{array}$

we then have $a|𝐯{|}^2-2𝐒𝐯-b=0$ where $𝐒=(S_x,S_y),$ 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 Template:Math at infinity) – ${|𝐯-\frac{𝐒}{a}|}^2=\frac{b}{a}+\frac{|𝐒{|}^2}{a^2},$ giving the circumcenter $\frac{𝐒}{a}$ and the circumradius $\sqrt{\frac{b}{a}+\frac{|𝐒{|}^2}{a^2}}.$ A similar approach allows one to deduce the equation of the circumsphere of a tetrahedron.

Parametric equation

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

$\hat{n}=\frac{(P_2-P_1)\times (P_3-P_1)}{|(P_2-P_1)\times (P_3-P_1)|}.$

Hence, given the radius, Template:Mvar, center, Template:Mvar, a point on the circle, Template:Math and a unit normal of the plane containing the circle, Template:Tmath one parametric equation of the circle starting from the point Template:Math and proceeding in a positively oriented (i.e., right-handed) sense about Template:Tmath is the following:

$\mathrm{R}(s)={\mathrm{P}}_{\mathrm{c}}+\cos \left(\frac{\mathrm{s}}{\mathrm{r}}\right)(P_0-P_c)+\sin \left(\frac{\mathrm{s}}{\mathrm{r}}\right)[\hat{n}\times (P_0-P_c)].$

Trilinear and barycentric coordinates

An equation for the circumcircle in trilinear coordinates Template:Math is $\frac{a}{x}+\frac{b}{y}+\frac{c}{z}=0.$ An equation for the circumcircle in barycentric coordinates Template:Math is $\frac{a^2}{x}+\frac{b^2}{y}+\frac{c^2}{z}=0.$

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

Higher dimensions

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

$\begin{array}{cc}𝐚& =𝐀-𝐂,\\ 𝐛& =𝐁-𝐂.\end{array}$

The circumradius Template:Mvar is then

$r=\frac{\Vert 𝐚\Vert \Vert 𝐛\Vert \Vert 𝐚-𝐛\Vert }{2\Vert 𝐚\times 𝐛\Vert }=\frac{\Vert 𝐚-𝐛\Vert }{2\sin \theta }=\frac{\Vert 𝐀-𝐁\Vert }{2\sin \theta },$

where Template:Mvar is the interior angle between Template:Math and Template:Math. The circumcenter, Template:Math, is given by

$p_0=\frac{({\Vert 𝐚\Vert }^2𝐛-{\Vert 𝐛\Vert }^2𝐚)\times (𝐚\times 𝐛)}{2{\Vert 𝐚\times 𝐛\Vert }^2}+𝐂.$

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:

$\begin{array}{cc}𝐮\times (𝐯\times 𝐰)& =(𝐮\cdot 𝐰)𝐯-(𝐮\cdot 𝐯)𝐰,\\ {\Vert 𝐮\times 𝐯\Vert }^2& ={\Vert 𝐮\Vert }^2{\Vert 𝐯\Vert }^2-(𝐮\cdot 𝐯{)}^2.\end{array}$

This gives us the following equation for the circumradius Template:Mvar:

$r=\frac{\Vert 𝐚\Vert \Vert 𝐛\Vert \Vert 𝐚-𝐛\Vert }{2\sqrt{{\Vert 𝐚\Vert }^2{\Vert 𝐛\Vert }^2-(𝐚\cdot 𝐛{)}^2}}$

and the following equation for the circumcenter Template:Math:

$p_0=\frac{(({\Vert 𝐚\Vert }^2𝐛-{\Vert 𝐛\Vert }^2𝐚)\cdot 𝐛)𝐚-(({\Vert 𝐚\Vert }^2𝐛-{\Vert 𝐛\Vert }^2𝐚)\cdot 𝐚)𝐛}{2({\Vert 𝐚\Vert }^2{\Vert 𝐛\Vert }^2-(𝐚\cdot 𝐛{)}^2)}+𝐂$

which can be simplified to:

$p_0=\frac{{\Vert 𝐚\Vert }^2{\Vert 𝐛\Vert }^2(𝐚+𝐛)-(𝐚\cdot 𝐛)({\Vert 𝐚\Vert }^2𝐛+{\Vert 𝐛\Vert }^2𝐚)}{2({\Vert 𝐚\Vert }^2{\Vert 𝐛\Vert }^2-(𝐚\cdot 𝐛{)}^2)}+𝐂$

Circumcenter coordinates

Cartesian coordinates

The Cartesian coordinates of the circumcenter $U=(U_x,U_y)$ are

$\begin{array}{cc}{U}_x& =\frac{1}{D}[(A_x^2+A_y^2)(B_y-C_y)+(B_x^2+B_y^2)(C_y-A_y)+(C_x^2+C_y^2)(A_y-B_y)]\\ U_y& =\frac{1}{D}[(A_x^2+A_y^2)(C_x-B_x)+(B_x^2+B_y^2)(A_x-C_x)+(C_x^2+C_y^2)(B_x-A_x)]\end{array}$

with

$D=2[A_x(B_y-C_y)+B_x(C_y-A_y)+C_x(A_y-B_y)].$

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

$\begin{array}{cc}U{\text{'}}_x& =\frac{1}{D^{\prime }}[C{\text{'}}_y({B{\text{'}}_x}^2+{B{\text{'}}_y}^2)-B{\text{'}}_y({C{\text{'}}_x}^2+{C{\text{'}}_y}^2)],\\ U{\text{'}}_y& =\frac{1}{D^{\prime }}[B{\text{'}}_x({C{\text{'}}_x}^2+{C{\text{'}}_y}^2)-C{\text{'}}_x({B{\text{'}}_x}^2+{B{\text{'}}_y}^2)]\end{array}$

with

$D^{\prime }=2(B{\text{'}}_{x}C{\text{'}}_y-B{\text{'}}_{y}C{\text{'}}_x).$

Due to the translation of vertex Template:Mvar to the origin, the circumradius Template:Mvar can be computed as

$r=\Vert U^{\prime }\Vert =\sqrt{{U{\text{'}}_x}^2+{U{\text{'}}_y}^2}$

and the actual circumcenter of Template:Math follows as

$U=U^{\prime }+A$

Trilinear coordinates

The circumcenter has trilinear coordinatesTemplate:Sfnp

$\cos \alpha :\cos \beta :\cos \gamma$

where Template:Mvar are the angles of the triangle.

In terms of the side lengths Template:Mvar, the trilinears are

$a(b^2+c^2-a^2):b(c^2+a^2-b^2):c(a^2+b^2-c^2).$

Barycentric coordinates

The circumcenter has barycentric coordinates

$a^2(b^2+c^2-a^2):b^2(c^2+a^2-b^2):c^2(a^2+b^2-c^2),$

where Template:Mvar are edge lengths Template:Mvar respectively) of the triangle.

In terms of the triangle's angles Template:Mvar, the barycentric coordinates of the circumcenter are

$\sin 2\alpha :\sin 2\beta :\sin 2\gamma .$

Circumcenter vector

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

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

Here Template:Mvar is the vector of the circumcenter and Template:Mvar are the vertex vectors. The divisor here equals Template:Math where Template:Mvar is the area of the triangle. As stated previously

$\begin{array}{cc}𝐚& =𝐀-𝐂,\\ 𝐛& =𝐁-𝐂.\end{array}$

Cartesian coordinates from cross- and dot-products

In Euclidean space, there is a unique circle passing through any given three non-collinear points Template:Math. 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

${\mathrm{P}}_{\mathrm{1}}=\left[\begin{array}{c}{x}_1\\ y_1\\ z_1\end{array}\right],{\mathrm{P}}_{\mathrm{2}}=\left[\begin{array}{c}{x}_2\\ y_2\\ z_2\end{array}\right],{\mathrm{P}}_{\mathrm{3}}=\left[\begin{array}{c}{x}_3\\ y_3\\ z_3\end{array}\right]$

Then the radius of the circle is given by

$\mathrm{r}=\frac{|P_1-P_2||P_2-P_3||P_3-P_1|}{2|(P_1-P_2)\times (P_2-P_3)|}$

The center of the circle is given by the linear combination

${\mathrm{P}}_{\mathrm{c}}=\alpha P_1+\beta P_2+\gamma P_3$

where

$\begin{array}{c}\alpha =\frac{{|P_2-P_3|}^2(P_1-P_2)\cdot (P_1-P_3)}{2{|(P_1-P_2)\times (P_2-P_3)|}^2}\\ \beta =\frac{{|P_1-P_3|}^2(P_2-P_1)\cdot (P_2-P_3)}{2{|(P_1-P_2)\times (P_2-P_3)|}^2}\\ \gamma =\frac{{|P_1-P_2|}^2(P_3-P_1)\cdot (P_3-P_2)}{2{|(P_1-P_2)\times (P_2-P_3)|}^2}\end{array}$

Location relative to the triangle

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.

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.

Angles

The angles which the circumscribed circle forms with the sides of the triangle coincide with angles at which sides meet each other. The side opposite angle Template:Math meets the circle twice: once at each end; in each case at angle Template:Math (similarly for the other two angles). This is due to the alternate segment theorem, which states that the angle between the tangent and chord equals the angle in the alternate segment.

Triangle centers on the circumcircle

In this section, the vertex angles are labeled Template:Mvar and all coordinates are trilinear coordinates:

• Steiner point: the non-vertex point of intersection of the circumcircle with the Steiner ellipse.

$\frac{bc}{b^2-c^2}:\frac{ca}{c^2-a^2}:\frac{ab}{a^2-b^2}$

(The Steiner ellipse, with center = centroid (Template:Mvar), is the ellipse of least area that passes through Template:Mvar. An equation for this ellipse is Template:Nowrap

• Tarry point: antipode of the Steiner point

$\sec (A+\omega ):\sec (B+\omega ):\sec (C+\omega )$

• Focus of the Kiepert parabola:

$\csc (B-C):\csc (C-A):\csc (A-B).$

Other properties

The diameter of the circumcircle, called the circumdiameter and equal to twice the circumradius, can be computed as the length of any side of the triangle divided by the sine of the opposite angle:

$\text{diameter}=\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}.$

As a consequence of the law of sines, it does not matter which side and opposite angle are taken: the result will be the same.

The diameter of the circumcircle can also be expressed as

$\begin{array}{cc}\text{diameter}& =\frac{abc}{2\cdot \text{area}}=\frac{|AB||BC||CA|}{2|\mathrm{\Delta }ABC|}\\ & =\frac{abc}{2\sqrt{s(s-a)(s-b)(s-c)}}\\ & =\frac{2abc}{\sqrt{(a+b+c)(-a+b+c)(a-b+c)(a+b-c)}}\end{array}$

where Template:Mvar are the lengths of the sides of the triangle and $s=\frac{a+b+c}{2}$ is the semiperimeter. The expression $\sqrt{s(s-a)(s-b)(s-c)}$ above is the area of the triangle, by Heron's formula. Trigonometric expressions for the diameter of the circumcircle include

$\text{diameter}=\sqrt{\frac{2\cdot \text{area}}{\sin A\sin B\sin C}}.$

The triangle's nine-point circle has half the diameter of the circumcircle.

In any given triangle, the circumcenter is always collinear with the centroid and orthocenter. The line that passes through all of them is known as the Euler line.

The isogonal conjugate of the circumcenter is the orthocenter.

The useful minimum bounding circle of three points is defined either by the circumcircle (where three points are on the minimum bounding circle) or by the two points of the longest side of the triangle (where the two points define a diameter of the circle). It is common to confuse the minimum bounding circle with the circumcircle.

The circumcircle of three collinear points is the line on which the three points lie, often referred to as a circle of infinite radius. Nearly collinear points often lead to numerical instability in computation of the circumcircle.

Circumcircles of triangles have an intimate relationship with the Delaunay triangulation of a set of points.

By Euler's theorem in geometry, the distance between the circumcenter Template:Mvar and the incenter Template:Mvar is

$\overline{OI}=\sqrt{R(R-2r)},$

where Template:Mvar is the incircle radius and Template:Mvar is the circumcircle radius; hence the circumradius is at least twice the inradius (Euler's triangle inequality), with equality only in the equilateral case.

The distance between Template:Mvar and the orthocenter Template:Mvar is

$\overline{OH}=\sqrt{R^2-8R^2\cos A\cos B\cos C}=\sqrt{9R^2-(a^2+b^2+c^2)}.$

For centroid Template:Mvar and nine-point center Template:Mvar we have

$\begin{array}{cc}\overline{IG}& <\overline{IO},\\ 2\overline{IN}& <\overline{IO},\\ \stackrel{2}{\stackrel{‾}{OI}}& =2R\cdot \overline{IN}.\end{array}$

The product of the incircle radius and the circumcircle radius of a triangle with sides Template:Mvar is

$rR=\frac{abc}{2(a+b+c)}.$

With circumradius Template:Mvar, sides Template:Mvar, and medians Template:Mvar, we have

$\begin{array}{cc}3\sqrt{3}R& \ge a+b+c\\ 9R^2& \ge a^2+b^2+c^2\\ \frac{27}{4}{R}^2& \ge m_a^2+m_b^2+m_c^2.\end{array}$

If median Template:Mvar, altitude Template:Mvar, and internal bisector Template:Mvar all emanate from the same vertex of a triangle with circumradius Template:Mvar, then

$4R^2{h}^2(t^2-h^2)=t^4(m^2-h^2).$

Carnot's theorem states that the sum of the distances from the circumcenter to the three sides equals the sum of the circumradius and the inradius.Template:Sfnp Here a segment's length is considered to be negative if and only if the segment lies entirely outside the triangle.

If a triangle has two particular circles as its circumcircle and incircle, there exist an infinite number of other triangles with the same circumcircle and incircle, with any point on the circumcircle as a vertex. (This is the Template:Math case of Poncelet's porism). A necessary and sufficient condition for such triangles to exist is the above equality $\overline{OI}=\sqrt{R(R-2r)}.$Template:Sfnp

Cyclic polygons

Cyclic quadrilaterals

Template:Main A set of points lying on the same circle are called concyclic, and a polygon whose vertices are concyclic is called a cyclic polygon. Every triangle is concyclic, but polygons with more than three sides are not in general.

Cyclic polygons, especially four-sided cyclic quadrilaterals, have various special properties. In particular, the opposite angles of a cyclic quadrilateral are supplementary angles (adding up to 180° or π radians).

See also

• Circumcenter of mass
• Circumscribed sphere
• Circumcevian triangle
• Inscribed circle
• Kosnita theorem
• Lester's theorem
• Problem of Apollonius

External links

• Derivation of formula for radius of circumcircle of triangle at Mathalino.com
• Semi-regular angle-gons and side-gons: respective generalizations of rectangles and rhombi at Dynamic Geometry Sketches, interactive dynamic geometry sketch.
• Weisstein, Eric W. "Circumcircle", "Cyclic Polygon". MathWorld.
• Triangle circumcircle and circumcenter With interactive animation
• An interactive Java applet for the circumcenter