Circumscribed circle

In geometry, the circumscribed circle or circumcircle of a polygon is a circle which passes through all the vertices of the polygon. The center of this circle is called the circumcenter.

A polygon which has a circumscribed circle is called a cyclic polygon. All regular simple polygons, all triangles and all rectangles are cyclic.

A related notion is the one of a minimum bounding circle, which is the smallest circle that completely contains the polygon within it. Not every polygon has a circumscribed circle, as the vertices of a polygon do not need to all lie on a circle, but every polygon has unique minimum bounding circle, which may be constructed by a linear time algorithm. Even if a polygon has a circumscribed circle, it may not coincide with its minimum bounding circle; for example, for an obtuse triangle, the minimum bounding circle has the longest side as diameter and does not pass through the opposite vertex.

Circumcircles of triangles


All triangles are cyclic, i.e. every triangle has a circumscribed circle.

The circumcenter of a triangle can be found as the intersection of the three perpendicular bisectors. (A perpendicular bisector is a line that forms a right angle with one of the triangle's sides and intersects that side at its midpoint.) This is because the circumcenter is equidistant from any pair of the triangle's points, and all points on the perpendicular bisectors are equidistant from those points of 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.

The circumcenter's position depends on the type of triangle:
 * If and only if a triangle is acute (all angles smaller than a right angle), the circumcenter lies inside the triangle
 * If and only if it is obtuse (has one angle bigger than a right angle), the circumcenter lies outside
 * If and only if it is a right triangle, the circumcenter lies on one of its sides (namely, the hypotenuse). This is one form of Thales' theorem.

Diameter
The diameter of the circumcircle can be computed as the length of any side of the triangle, divided by the sine of the opposite angle. (As a consequence of the law of sines, it doesn't matter which side is taken: the result will be the same.) The triangle's nine-point circle has half the diameter of the circumcircle. The diameter of the circumcircle of the triangle &Delta;ABC is


 * $$D = \frac{abc}{2\cdot\text{area}} = \frac{|AB| |BC| |CA|}{2|\Delta ABC|}= \frac{abc}{2\sqrt{s(s-a)(s-b)(s-c)}}= \frac{2abc}{\sqrt{(a+b+c)(-a+b+c)(a-b+c)(a+b-c)}}= \frac{abc}{\sqrt{\frac{(a^2+b^2+c^2)^2}{4}-\frac{(a^4+b^4+c^4)}{2}}}= s(2-\frac{4}{n})tan(\frac{180}{n})$$

where a, b, c are the lengths of the sides of the triangle and $$s= \frac{(a + b + c)}{2}$$ is the semiperimeter. The radical in the second denominator above is the area of the triangle, by Heron's formula.

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.

Radius
The circumcircle of three collinear points is the line on which the 3 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.

The radius is:


 * $$R= \frac{abc}{4A}= \frac{abc}{\sqrt{(a^2+b^2+c^2)^2-2(a^4+b^4+c^4)}}$$


 * $$R= s(1-\frac{2}{n})tan(\frac{180}{n})$$

For a right triangle:


 * $$R= \frac{\sqrt{a^2+b^2}}{2}= \frac{c}{2}$$

Area
The area is:
 * $$A= \frac{a^2b^2c^2}{16A^2}\pi= \frac{a^2b^2c^2}{(a^2+b^2+c^2)^2-2(a^4 + b^4 +c^4)}\pi= s^2\frac{\pi}{4}= s^2\pi(1-\frac{2}{n})^2tan^2(\frac{180}{n})$$

Perimeter
The perimeter is:
 * $$P= \frac{abc}{2A}\pi= \frac{abc}{\sqrt{\frac{(a^2+b^2+c^2)^2}{4}-\frac{(a^4+b^4+c^4)}{2}}}\pi= s\pi= s\pi(2-\frac{4}{n})tan(\frac{180}{n})$$

Circumcircles of quadrilaterals
The Radius is $$R= \sqrt{\frac{ab(c^3d+cd^3)+a^2b^2(c^2+d^2)+cd(a^3b+ab^3)+c^2d^2(a^2+b^2)}{16A^2}}= \frac{\sqrt{l^2+w^2}}{2}$$

Circumcircles of polygons
The circumradius of a regular n-sided polygon is:
 * $$R= \frac{s}{2sin(\frac{180}{n})}$$

For the diameter:
 * $$D= s\frac{2}{2sin(\frac{180}{n})}$$

Area and Perimeter
Circmcircle area:
 * $$A= s^2\frac{\pi}{4sin^2(\frac{180}{n})}$$

Circmcircle perimeter:
 * $$P= s\frac{2\pi}{2sin(\frac{180}{n})}$$

Circumcircle equations
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. Thus suppose that


 * $$\mathbf{A} = (A_x,A_y)$$
 * $$\mathbf{B} = (B_x,B_y)$$
 * $$\mathbf{C} = (C_x,C_y)$$

are the coordinates of points A, B, and C. The circumcircle is then the locus of points v = (vx,vy) in the Cartesian plane satisfying the equations


 * $$|\mathbf{v}-\mathbf{u}|^2 - r^2 = 0$$
 * $$|\mathbf{A}-\mathbf{u}|^2 - r^2 = 0$$
 * $$|\mathbf{B}-\mathbf{u}|^2 - r^2 = 0$$
 * $$|\mathbf{C}-\mathbf{u}|^2 - r^2 = 0$$

guaranteeing that the points A, B, v are all the same distance r2 from the common center u of the circle. Using the polarization identity, these equations reduce to a the condition that the matrix


 * $$\begin{vmatrix}

\end{vmatrix}$$
 * \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

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


 * $$\det\begin{vmatrix}

\end{vmatrix}=0$$
 * \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

Expanding by cofactor expansion, let


 * $$\quad

S_x=\frac{1}{2}\det\begin{vmatrix} \end{vmatrix},\quad S_y=\frac{1}{2}\det\begin{vmatrix} A_x & |\mathbf{A}|^2 & 1 \\ B_x & |\mathbf{B}|^2 & 1 \\ C_x & |\mathbf{C}|^2 & 1 \end{vmatrix},$$
 * \mathbf{A}|^2 & A_y & 1 \\
 * \mathbf{B}|^2 & B_y & 1 \\
 * \mathbf{C}|^2 & C_y & 1
 * $$a=\det\begin{vmatrix}

A_x & A_y & 1 \\ B_x & B_y & 1 \\ C_x & C_y & 1 \end{vmatrix},\quad b=\det\begin{vmatrix} A_x & A_y & |\mathbf{A}|^2 \\ B_x & B_y & |\mathbf{B}|^2 \\ C_x & C_y & |\mathbf{C}|^2 \end{vmatrix}$$ we then have a|v|2 &minus; 2Sv &minus; b = 0 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), |v &minus; S/a|S/a2, giving the circumcenter S/a and the circumradius √ (b/a + |S|2/a2). A similar approach allows one to deduce the equation of the circumsphere of a tetrahedron.

An equation for the circumcircle in trilinear coordinates x : y : z is a/x + b/y + c/z = 0. An equation for the circumcircle in barycentric coordinates x : y : z is 1/x + 1/y + 1/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.

Cartesian coordinates
The Cartesian coordinates of the circumcenter are

$$ (( B_yA_x^2  - C_yA_x^2  - B_y^2A_y  + C_y^2A_y + B_x^2C_y + A_y^2B_y  + C_x^2A_y  - C_y^2B_y - C_x^2B_y  - B_x^2A_y  + B_y^2C_y  - A_y^2C_y ) / D, $$

$$( A_x^2C_x  + A_y^2C_x  + B_x^2A_x  - B_x^2C_x  + B_y^2A_x  - B_y^2C_x  - A_x^2B_x  - A_y^2B_x  - C_x^2A_x  + C_x^2B_x  - C_y^2A_x  + C_y^2B_x)  / D) $$

with


 * $$ D = 2( A_yC_x + B_yA_x - B_yC_x - A_yB_x -C_yA_x + C_yB_x ).\, $$

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 $$ A' = A - A = (A'_x,A'_y) = (0,0)$$. In this case, the coordinates of the vertices B' = B &minus; A and C' = C &minus; A represent the vectors from vertex A' to these vertices. Observe that this trivial translation is possible for all triangles and the circumcenter coordinates of the triangle A'B'C' follow as


 * $$ (( C'_y(B^{'2}_x + B^{'2}_y) - B'_y(C^{'2}_x+C^{'2}_y) )/ D', \, $$


 * $$( B'_x(C^{'2}_x+C^{'2}_y) - C'_x(B^{'2}_x+B^{'2}_y) )/ D') \, $$

with


 * $$ D' = 2( B'_xC'_y - B'_yC'_x ). \, $$

Barycentric coordinates as a function of the side lengths
The circumcenter has trilinear coordinates (cos &alpha;, cos &beta;, cos &gamma;) where &alpha;, &beta;, &gamma; are the angles of the triangle. The circumcenter has barycentric coordinates


 * $$ \left( a^2(-a^2 + b^2 + c^2), \;b^2(a^2 - b^2 + c^2), \;c^2(a^2 + b^2 - c^2) \right), \, $$

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

Barycentric coordinates from cross- and dot-products
In Euclidean space, there is a unique circle passing through any given three non-collinear points P1, P2, and 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



\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



\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



\mathrm{P_c} = \alpha \, P_1 + \beta \, P_2 + \gamma \, P_3 $$

where



\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} $$

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



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

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



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

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

Triangle centers on the circumcircle of triangle ABC
In this section, the vertex angles are labeled A, B, C and all coordinates are trilinear coordinates:


 * Steiner point = $$\frac{bc}{b^2-c^2} : \frac{ca}{c^2-a^2} : \frac{ab}{a^2-b^2}$$ = the nonvertex point of intersection of the circumcircle with the Steiner ellipse. (The Steiner ellipse, with center = centroid(ABC), is the ellipse of least area that passes through A, B, and C.  An equation for this ellipse is 1/(ax) + 1/(by) + 1/(cz) = 0.)


 * Tarry point = sec (A + ω) : sec (B + ω) : sec (C + ω) = antipode of the Steiner point


 * Focus of the Kiepert parabola = csc (B &minus; C) : csc (C &minus; A) : csc (A &minus; B)

Cyclic quadrilaterals


Quadrilaterals that can be circumscribed have particular properties including the fact that opposite angles are supplementary angles (adding up to 180&deg; or π radians).

Interactive

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

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