Come on friends, today we will know about the circumscribed circle. In geometry , the circumscribed circle or circumcircle of a polygon is a circle that passes through all the vertices of the polygon. The center of this circle is called the circumcentre and its radius is called the circumradius . Not every polygon has a bounded circle. A polygon is one that is called a cyclic polygon , or sometimes a unicyclic polygon because its vertices are concentric . All triangles , all regular simple polygons , all rectangles , all isosceles trapeziums , and all right-sided kites are cyclic. Let’s start about the circumscribed circle

A related notion is that of the minimum bounding circle , which is the smallest circle that contains a polygon as a whole, if the center of the circle is within the polygon. Each polygon has a unique minimum bounding circle, which can be constructed by a linear time algorithm. [1] Even if the polygon has an enclosed circle, it may be different from its minimum boundary circle. For example, for an obtuse triangle , the minimum bounding circle has the longest side as the diameter and does not pass through the opposite vertex.

**Triangle**

All triangles are cyclic; That is, every triangle has a circumscribed circle.

**Upright and compass construction**

(a triangle with an angle greater than a right angle)

The circumcentre 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 circumcentre is the point from which they cross. Any point on the bisector is equidistant from the two points it bisects, so that it follows that this point, at both bisectors, is equidistant from all three triangle vertices. The circumradius is the distance of any one of the three vertices from it.

**Alternative construction**

An alternative way to determine the circumcentre is to draw any two lines departing from each vertex at an angle of the common side, the common angle of departure minus the angle of the opposite vertex is 90°. (Drawing a line at a negative angle means going outside the triangle, in case the opposite angle is obtuse.)

In coastal navigation , the circumcircle of a triangle is sometimes used as a way of obtaining a position line using a sextant , when no compass is available. The horizontal angle between two points defines the circumference on which the observer is located. (Circumscribed Circle)

**Circular equation**

**Cartesian coordinates**

In the Euclidean plane , it is possible to explicitly give an equation of the circumcircle in terms of the Cartesian coordinates of the vertices of the present triangle. it looks as though (Circumscribed Circle)

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

The coordinates of points *A* , *B* , and *C are . ***Then the locus of the points v** = ( *v *_{x} , *v *_{y} ) in the Cartesian plane satisfying the circumscribed equations is.

{\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}}}

This guarantees that the points A , B , C , and v are all equidistant r from the common center u of the circle. Using polarization detection , 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}}}

There is a non-zero kernel . Thus the circumcircle can alternatively be described as the zero locus 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 co- operative 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}}}

Then we have one. **V** | ^{2} − 2 **Sv** – *b* = 0 And, assuming that the three points were not in a line (otherwise the circumcircle is the line which can also be seen as a generalized circle with S at infinity), | **V** – **S** / *A* | ^{2} = *b* / *a* + | **S** | ^{2} / *a *^{2 , }*giving* circumcenter **s** / *a* and circumradius b / *a* + | **S** | ^{2} /*one *^{2} . A similar approachallows one to derive the equation for the perimeter tetrahedron

**Parametric equation**

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})|}}.}

Therefore, given the radius, r , center, P c , a point on the circle, P 0 and a unit normal of the plane of the circle, , a parametric equation of a circle starting at the point P 0 and oriented positively (i.e. , right hand ) proceed about any of the following:

{\textstyle {\widehat {n}}}{\displaystyle \scriptstyle {\widehat {n}}}

{\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].}

**Trilinear and barycentric coordinates**

An equation in the circumcircle for three-linear coordinates *x* : *y* : *z* is *a* / *x* + *b* / *y* + *c* / *z* = 0 . The equation for an circumcircle in barycentric coordinates *x* : *y* : *z* is *a *^{2} / *x* + *b *^{2} / *y* + *c *^{2} / *z* = 0 . The isogonal conjugate of the circumcircle at infinity is the line, given in three-linear coordinates by ax + by + cz = 0 and by in barycentric coordinates x + y + z = 0 .

**High dimensions**

Additionally, the circumcircle of a triangle embedded in *d* dimensions can be found using a generalized method. Let **a** , **b** , and **c** be *d* dimensional points, which form the vertices of a triangle. We begin by moving the system to place **C in the original position:**

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

Radius, *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 is the interior *angle* between **a** and **b . **The circumcentre, *p *_{0} , 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 because the cross product is not defined in other dimensions, but can be generalized to other dimensions by replacing the cross products with the following identities:

{\displaystyle {\begin{aligned}(\mathbf {a} \times \mathbf {b} )\times \mathbf {c} &=(\mathbf {a} \cdot \mathbf {c} )\mathbf {b} -(\mathbf {b} \cdot \mathbf {c} )\mathbf {a} ,\\\mathbf {a} \times (\mathbf {b} \times \mathbf {c} )&=(\mathbf {a} \cdot \mathbf {c} )\mathbf {b} -(\mathbf {a} \cdot \mathbf {b} )\mathbf {c} ,\\\left\|\mathbf {a} \times \mathbf {b} \right\|^{2}&=\left\|\mathbf {a} \right\|^{2}\left\|\mathbf {b} \right\|^{2}-(\mathbf {a} \cdot \mathbf {b} )^{2}.\end{aligned}}}

**Circumcentre coordinates**

**Cartesian coordinates**

Cartesian coordinates are of circumcenter

{\displaystyle U=\left(U_{x},U_{y}\right)}

{\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

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

The translation of this vertex without loss of generality can be expressed in a simple form after *the* Cartesian coordinate system of a to the origin, ie, when *a* ‘ = *a* – *a* = ( *a* ‘ _{x} , *a* ‘ _{y} ) = (0, 0) . In this case, the coordinates of the vertices *B* = *B* – *A* and *C* = *C* – *A represent the vectors from the vertex A* to these vertices. Note that this trivial translation is possible for all triangles and the circumcentre of triangle *A*

{\displaystyle U'=(U'_{x},U'_{y})}

Follow *B C* _

{\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 at the origin of vertex *A* , the perimeter *R* can be calculated as

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

and the real circumcentre of *ABC* is

{\displaystyle U = U '+ A}

**Triangular coordinates**

The circumcentre has three linear coordinates [3]

Because α : Because β : Because

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

With respect to the lengths of the sides a, b, c , are trilinear [4]

{\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).}

**Barycentric coordinates**

The circumcentre has barycentric coordinates

{\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 the lengths of the sides of the triangle ( *BC* , *CA* , *AB* respectively).

The barycentric coordinates of the circumcentre with respect to the angles of the triangle are

\alpha ,\beta ,\gamma ,

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

**circumference vectorcent**

Since the Cartesian coordinates of any point are the weighted average of the vertices, with the weights being the barycentric coordinates of the point normalized to the sum of unity, the perimeter vector can be written as (Circumscribed Circle)

{\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 circumcentre vector and *A, B, C* are the vertex vectors. Here the denominator is equal to 16 *S*^{2 }where *S* is the area of the triangle. as stated before

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

**Cartesian coordinates from cross- and dot-products**

In Euclidean space, there is an exclusive cycle passing through any three non-collinear points *P *_{1} , *P *_{2} , and *P *_{3} . It is possible to use Cartesian coordinates to represent these points as spatial vectors, use the dot product and cross product to calculate the radius and center of the circle. Army

{\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

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

Where from

{\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}}}

**Relative position of the triangle**

The position of the perimeter depends on the type of triangle:

- For an acute triangle (all angles less than a right angle) the circumcentre is always inside the triangle.
- For a right angled triangle, the circumcentre always lies at the mid-point of the hypotenuse. It is a variant of Thales’ theorem.
- For an obtuse triangle (a triangle with an angle greater than a right angle) the circumcentre always lies outside the triangle.

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

**Angles**

The angles forming a circumscribed circle with the sides of a triangle correspond to the angles at which the sides meet. The opposite angle *α* meets the circle twice: once at each end; At angle *α* in each case (similarly for the other two angles). This is due to **the alternate segment theorem** , which states that the angle between the tangent and the chord is equal to the angle in the alternate segment.

**Triangle center on circumcircle of triangle ABC**

*In this section, the* vertex angles are labeled *A* , *B* , *C* and all coordinates are trilinear coordinates:

- Steiner point =
*BC*/(*B2*–^{C2}):*CA*/(*C2**–*^{A2}):*AB*/(*A2 – B2*^{)}=^{non}–^{vertex }*point*of intersection of the circumference with the Steiner*ellipse*^{. }(The Steiner ellipse, with center = centroid (*abc*), is an ellipse of least area passing through*a*,*b*, and*c*. An equation for this ellipse is*1*/(*ax*) + 1/(*by*) + 1/ (^{}^{}^{}^{}*cz*) =*0.**,* - Terri point = sec (
*a*+ ): sec (*b*+ ): sec (*c*+ ) = antipode of Steiner point - Keynote focus of the parabola = csc (
*B*–*C*): csc (*C*–*A*): csc (*A*–*B*).

**Other properties**

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

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

As a consequence of the law of sine, no matter which side and the opposite angle is taken: the result will be the same.

The diameter of the circumcircle can also be expressed as

{\displaystyle {\begin{aligned}{\text{diameter}}&{}={\frac {abc}{2\cdot {\text{area}}}}={\frac {|AB||BC||CA|}{2|\Delta ABC|}}\\[5pt]&{}={\frac {abc}{2{\sqrt {s(s-a)(s-b)(s-c)}}}}\\[5pt]&{}={\frac {2abc}{\sqrt {(a+b+c)(-a+b+c)(a-b+c)(a+b-c)}}}\end{aligned}}}

where *a* , *b* , *c* are the lengths of the sides of the triangle and *s* = ( *a* + *b* + *c* )/2 is the semiperimeter. The area of the triangle above is given by the formula of Izhar Heron. ^{[6]} Trigonometric expressions for the diameter of the circumference include

{\sqrt {\scriptstyle {s(s-a)(s-b)(s-c)}}}

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

The nine-pointed circle of a triangle has half the diameter of the circle.

In a given triangle, the circumcentre is always collinear with the centroid and the orthocentre. The line passing through all of them is called Euler line.

The isogonal conjugate is the circumcenter of the orthocenter.

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

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 the calculation of the circle. The circles of triangles are closely related to the Delaunay triangle of a set of points.

By Euler’s theorem in geometry, the distance between the circumcenter *O* and the incenter *I* is

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

where *r* is the incircle radius and *R* is the circumcircle radius; Therefore the perimeter is at least twice the inradius (Euler’s triangle inequality), with equality only in the equilateral case. ^{[8] }^{[9]}^{}^{}

The distance between *O* and orthocenter *H is *

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

For the centroid *G* and the nine-point center *N* we have

*IG < IO*

*2IN < IO*

*OI ^{2} = 2R . IN*

The product of the incircle radii and the circumcircle radii of a triangle with sides *a* , *b* , and *c* is

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

With perimeter *R* , sides *a* , *b* , *c* , and medians *m *_{a} , *m *_{b} , and *m *_{c} , we have

{\displaystyle {\begin{aligned}3{\sqrt {3}}R&\geq a+b+c\\[5pt]9R^{2}&\geq a^{2}+b^{2}+c^{2}\\[5pt]{\frac {27}{4}}R^{2}&\geq m_{a}^{2}+m_{b}^{2}+m_{c}^{2}.\end{aligned}}}

If the median *m* , height *h* , and internal bisector *t* all originate from the same vertex of a triangle of perimeter *R , then*

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

The Carnot theorem states that for three sides the sum of the distances from the circumcenter equals the sum of the circumradius and the inradius. ^{[15]} Here the length of a segment is considered negative if and only if the segment lies entirely outside the triangle.

If a triangle consists of two particular circles such as its circumcircle and incircle, there exist an infinite number of other triangles with the same circumcircle and circle as a vertex that has any point on the circumcircle. *(This is the n* = 3 case of Ponslet’s hole ). A necessary and sufficient condition for the existence of such triangles is the above equality

OI = {\sqrt {R (R-2r)}}.

**Cyclic quadrilateral**

Quadrilaterals that can be bounded have special properties, including the fact that opposite angles are complementary angles (adding up to 180° or radians).

For a cyclic polygon with an odd number of sides, all angles are equal if and only if the polygon is regular. In a cyclic polygon that has an even number of sides, all angles are equal if and only if the alternate sides are equal (that is, the sides 1, 3, 5, … are equal, and the sides 2, 4, 6, … are equal). ^{[17]}

A cyclic pentagon with rational sides and area is known as a Robbins pentagon; In all known cases, its diagonals also have rational lengths.

In any cyclic *n* -gon with even *n, the* sum of one set of alternate angles (the first, third, fifth, etc.) is equal to the sum of the second set of alternate angles. This can be proved by induction from the *n* = 4 case, replacing one side with three more sides in each case and noting that these three new sides together with the old side form a quadrilateral which itself has this property. it occurs; The alternate angles of the subsequent quadrilateral denote the sum of the alternate angles of the previous *n -gon.*

Let one *n* -gon be inscribed in a circle, and let the second *n* -gon be tangents to that circle at the vertices of the first *n -gon. *Then the product of the distances from any point *P* on the circle perpendicular to the sides of the first *n* – gon from *P is equal to the product of the perpendicular distances from P* to the sides of the second *n* – gon.

**Point on the circumcircle**

Let the vertices on the unit circle in _{a cyclic }*n -gon be A *_{1} , …, *A *_{n} . Then for any point *M on the* minor arc *A *_{1 }*A *_{n} , the distance from *M* to the vertices satisfies

For a regular *n* -gon, if the distances from any point are the circumcircle to the vertices , then *MA _{i}*

{\displaystyle 3(MA_{1}^{2}+MA_{2}^{2}+...+MA_{n}^{2})^{2}=2n(MA_{1}^{4) }+MA_{2}^{4}+...+MA_{n}^{4}).}

**Polygon rotation constant**

Any regular polygon is cyclic. Consider a unit circle, then circumscribe a regular triangle such that each side touches the circle. Circle a circle, then circle a square. Circle a circle again, then circle a regular 5-gon, and so on. The radii of the circumscribed circles converge *in the so called polygon which operates the constant*

{\displaystyle \prod _{n=3}^{\infty }{\frac {1}{\cos \left({\frac {\pi }{n}}\right)}}=8.7000366\ldots .}

(Sequence A051762 in OEIS ). The inverse of this constant is the Kepler–Bauwkamp constant.