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| [[File:Circumscribed Polygon.svg|thumb|Circumscribed circle, ''C'', and circumcenter, ''O'', of a ''cyclic polygon'', ''P'']]
| | Eat a low fat, high fiber diet contain plenty of green vegetables and fresh fruits to reduce cellulite problem. Cellulite is evident by seeing skin dimpling and nodularity which is found in the pelvic region, lower limbs and abdomen. There's just no way around it, if you're serious about treating your cellulites you should go sign up for a gym membership. When massaging is carried out properly, it stimulates blood circulation in the cellulite area. Ongoing month-to-month maintenance classes are typically determined by the treating practitioner. <br><br>I use it for like three months now and I never had the smallest problem with it (like my hands slipping from the massager like it happened with other products);- A very good thing about it is also the fact that you can set the strength of the massage, air suction and heating. This is quite easy and simple for women who gave birth in the normal way. Bend your knees and keep your legs a few inches off the ground for a few seconds. And, once you finally achieve that sculpted figure you so badly desire, maintaining it will be easier than you could have ever thought possible. It is a change in the skin that occurs in most females and appears as dimpling or a 'cottage cheese' look in areas generally from the stomach down. <br><br>Lack of exercise, bad dietary habits, and poor circulation are equally to be blamed for this undesirable condition. It is in definite areas and begins on the inside surface of the thighs and the knees. Most of the sexual problems can easily preventable with homeopathic medicines and with self take care about vagina. Luckily, that desire can be completely fulfilled with cellulite exercises. Normal water is beneficial because it improves metabolic rate and it's also good for the epidermis. <br><br>But, the revolutionary new cellulite treatment called Cellulaze has swiftly established itself as the only answer that may remove the vast majority of evident cellulite. It is important for you to consider all the facts before deciding if this particular method of Denver plastic surgery is right for you. This is the right path that will lead you to a cellulite free body that you have dreamed of. However, losing to much body fluids can cause serious problems, and even death. Instead, focus on arming yourself with knowledge, and the real facts about cellulite. <br><br>So that was the answer to 'what are the benefits of Cellulaze cellulite treatment'. There are several things that appear to have been positively affective:. cellulite lotions are readily available in the market. They are utilized to boost eyesight through LASIK surgery, remove tattoos, as anti-aging treatments to go out of young looking skin, and of course cellulite removal. It has been suggested that the collagen in fibrous connective tissue (which tethers the skin to the underlying tissue) may be a major factor in why some ladies suffer while others don't.<br><br>If you have any questions about where and also the way to employ [http://www.samfinds.info/sitemap/ best cellulite cream 2014], you can contact us with the webpage. |
| In [[geometry]], the '''circumscribed circle''' or '''circumcircle''' of a [[polygon]] is a [[circle]] which passes through all the vertices of the polygon. The [[center (geometry)|center]] of this circle is called the '''circumcenter''' and its radius is called the '''circumradius'''.
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| A polygon which has a circumscribed circle is called a '''cyclic polygon''' (sometimes a '''concyclic polygon''', because the vertices are [[concyclic]]). All [[Regular polygon|regular]] [[simple polygon]]s, [[isosceles trapezoid]]s, all [[triangle]]s and all [[rectangle]]s are cyclic.
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| A related notion is the one of a [[Smallest circle problem|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 a unique minimum bounding circle, which may be constructed by a [[linear time]] algorithm.{{ref|Megiddo}} 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.
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| ==Triangles==
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| [[File:Circumcenter Construction.svg|right|565px|Construction of the circumcircle (red) and the circumcenter (red dot)]]
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| All triangles are cyclic, i.e. every triangle has a circumscribed circle.<ref group="nb">This can be proven on the grounds that the general equation for a circle with center (''a'', ''b'') and radius ''r'' in the [[Cartesian coordinate system]] is
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| :<math>\left(x - a \right)^2 + \left( y - b \right)^2=r^2.</math>
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| Since this equation has three parameters (''a'', ''b'', ''r'') only three pairs of points are required to determine the equation of a circle. Since a triangle is defined by its three vertices, and exactly three points are required to determine a circle, every triangle can be circumscribed.</ref>
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| The circumcenter of a triangle can be found as the intersection of any two of the three [[Bisection#Perpendicular bisectors|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 vertices, and all points on the perpendicular bisectors are equidistant from two of the vertices of the triangle.
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| [[File:Triangle circumcenter alternate construction.png|left|thumb|250px|Alternate construction of the circumcenter (intersection of broken lines)]]
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| An alternate method to determine the circumcenter is to draw any two lines each departing 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.)
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| In [[pilotage|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.
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| The circumcenter's position depends on the type of triangle:
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| *[[If and only if]] a triangle is acute (all angles smaller than a right angle), the circumcenter lies inside the triangle.
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| *If and only if it is obtuse (has one angle bigger than a right angle), the circumcenter lies outside the triangle.
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| *If and only if it is a right triangle, the circumcenter lies at the center of the [[hypotenuse]]. This is one form of [[Thales' theorem]].
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| <gallery>
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| Image:Triangle (Acute) Circumscribed.svg|The circumcenter of an acute triangle is inside the triangle
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| Image:Triangle (Right) Circumscribed.svg|The circumcenter of a right triangle is at the center of the hypotenuse
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| Image:Triangle (Obtuse) Circumscribed.svg|The circumcenter of an obtuse triangle is outside the triangle
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| </gallery>
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| 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 does not 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 Δ''ABC'' is
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| :<math>\begin{align}
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| \text{diameter} & {} = \frac{abc}{2\cdot\text{area}} = \frac{|AB| |BC| |CA|}{2|\Delta ABC|} \\
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| & {} = \frac{abc}{2\sqrt{s(s-a)(s-b)(s-c)}}\\
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| & {} = \frac{2abc}{\sqrt{(a+b+c)(-a+b+c)(a-b+c)(a+b-c)}}
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| \end{align}</math>
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| where ''a'', ''b'', ''c'' are the lengths of the sides of the triangle and {{nowrap|''s'' {{=}} (''a'' + ''b'' + ''c'')/2}} is the semiperimeter. The expression <math>\sqrt{\scriptstyle{s(s-a)(s-b)(s-c)}}</math> above is the area of the triangle, by [[Heron's formula]].{{ref|Coxeter}} Trigometric expressions for the diameter of the circumcircle include<ref>Dörrie, Heinrich, ''100 Great Problems of Elementary Mathematics'', Dover, 1965.</ref>{{rp|p.379}}
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| :<math>\text{diameter} = \sqrt{\frac{2 \cdot \text{area}}{\sin A \sin B \sin C}}.</math>
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| 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]].
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| The [[isogonal conjugate]] of the circumcenter is the [[orthocenter]].
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| The useful [[Smallest circle problem|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.
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| 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.
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| Circumcircles of triangles have an intimate relationship with the [[Delaunay triangulation]] of a [[Set (mathematics)|set]] of points.
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| ===Circumcircle equations===
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| 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
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| :<math>\mathbf{A} = (A_x,A_y)</math>
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| :<math>\mathbf{B} = (B_x,B_y)</math>
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| :<math>\mathbf{C} = (C_x,C_y)</math>
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| are the coordinates of points ''A'', ''B'', and ''C''. The circumcircle is then the locus of points '''v''' = (''v''<sub>x</sub>,''v''<sub>y</sub>) in the Cartesian plane satisfying the equations
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| :<math>|\mathbf{v}-\mathbf{u}|^2 = r^2</math>
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| :<math>|\mathbf{A}-\mathbf{u}|^2 = r^2</math>
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| :<math>|\mathbf{B}-\mathbf{u}|^2 = r^2</math>
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| :<math>|\mathbf{C}-\mathbf{u}|^2 = r^2</math>
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| guaranteeing that the points '''A''', '''B''', '''C''', and '''v''' are all the same distance ''r'' from the common center ''u'' of the circle. Using the [[polarization identity]], these equations reduce to the condition that the [[matrix (mathematics)|matrix]]
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| :<math>\begin{bmatrix}
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| |\mathbf{v}|^2 & -2v_x & -2v_y & -1 \\
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| |\mathbf{A}|^2 & -2A_x & -2A_y & -1 \\
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| |\mathbf{B}|^2 & -2B_x & -2B_y & -1 \\
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| |\mathbf{C}|^2 & -2C_x & -2C_y & -1
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| \end{bmatrix}</math>
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| has a nonzero [[kernel (linear algebra)|kernel]]. Thus the circumcircle may alternatively be described as the locus of zeros of the [[determinant]] of this matrix:
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| :<math>\det\begin{bmatrix}
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| |\mathbf{v}|^2 & v_x & v_y & 1 \\
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| |\mathbf{A}|^2 & A_x & A_y & 1 \\
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| |\mathbf{B}|^2 & B_x & B_y & 1 \\
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| |\mathbf{C}|^2 & C_x & C_y & 1
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| \end{bmatrix}=0.</math>
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| Expanding by [[cofactor expansion]], let
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| :<math>\quad
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| S_x=\frac{1}{2}\det\begin{bmatrix}
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| |\mathbf{A}|^2 & A_y & 1 \\
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| |\mathbf{B}|^2 & B_y & 1 \\
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| |\mathbf{C}|^2 & C_y & 1
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| \end{bmatrix},\quad
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| S_y=\frac{1}{2}\det\begin{bmatrix}
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| A_x & |\mathbf{A}|^2 & 1 \\
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| B_x & |\mathbf{B}|^2 & 1 \\
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| C_x & |\mathbf{C}|^2 & 1
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| \end{bmatrix},</math>
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| :<math>a=\det\begin{bmatrix}
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| A_x & A_y & 1 \\
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| B_x & B_y & 1 \\
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| C_x & C_y & 1
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| \end{bmatrix},\quad
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| b=\det\begin{bmatrix}
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| A_x & A_y & |\mathbf{A}|^2 \\
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| B_x & B_y & |\mathbf{B}|^2 \\
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| C_x & C_y & |\mathbf{C}|^2
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| \end{bmatrix}</math>
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| we then have a|'''v'''|<sup>2</sup> − 2'''Sv''' − ''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), {{nowrap|{{!}}'''v''' − '''S'''/''a''{{!}}<sup>2</sub> {{=}} ''b''/''a'' + {{!}}'''S'''{{!}}<sup>2</sup>/''a''<sup>2</sup>}}, giving the circumcenter '''S'''/''a'' and the circumradius {{nowrap|√(''b''/''a'' + {{!}}'''S'''{{!}}<sup>2</sup>/''a''<sup>2</sup>).}} A similar approach allows one to deduce the equation of the [[circumsphere]] of a [[tetrahedron]].
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| An equation for the circumcircle in [[trilinear coordinates]] ''x'' : ''y'' : ''z'' is {{nowrap|''a''/''x'' + ''b''/''y'' + ''c''/''z'' {{=}} 0}}. An equation for the circumcircle in [[barycentric coordinates (mathematics)|barycentric coordinates]] ''x'' : ''y'' : ''z'' is {{nowrap|''a''<sup>2</sup>/''x'' + ''b''<sup>2</sup>/''y'' + ''c''<sup>2</sup>/''z'' {{=}} 0}}.
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| The [[isogonal conjugate]] of the circumcircle is the line at infinity, given in [[trilinear coordinates]] by {{nowrap|''ax'' + ''by'' + ''cz'' {{=}} 0}} and in barycentric coordinates by {{nowrap|''x'' + ''y'' + ''z'' {{=}} 0}}.
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| 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 start by transposing the system to place '''C''' at the origin:
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| :<math>\mathbf{a} = \mathbf{A}-\mathbf{C},</math>
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| :<math>\mathbf{b} = \mathbf{B}-\mathbf{C}.</math>
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| The circumradius, ''r'', is then
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| :<math>r = \frac{\left\|\mathbf{a}\right\|\left\|\mathbf{b}\right\|\left\|\mathbf{a}-\mathbf{b}\right\|}
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| {2 \left\|\mathbf{a}\times\mathbf{b}\right\|}
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| = \frac{\left\|\mathbf{a}-\mathbf{b}\right\|}{2 \sin\theta}
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| = \frac{\left\|\mathbf{A}-\mathbf{B}\right\|}{2 \sin\theta},</math>
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| where ''θ'' is the interior angle between '''a''' and '''b'''. The circumcenter, ''p''<sub>0</sub>, is given by
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| :<math>p_0 = \frac{(\left\|\mathbf{a}\right\|^2\mathbf{b}-\left\|\mathbf{b}\right\|^2\mathbf{a})
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| \times (\mathbf{a} \times \mathbf{b})}
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| {2 \left\|\mathbf{a}\times\mathbf{b}\right\|^2} + \mathbf{C}.</math>
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| 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:
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| :<math>(\mathbf{a}\times\mathbf{b})\times\mathbf{c}=(\mathbf{a} \cdot \mathbf{c})\mathbf{b}-(\mathbf{b} \cdot \mathbf{c})\mathbf{a},</math>
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| :<math>\mathbf{a}\times(\mathbf{b}\times\mathbf{c})=(\mathbf{a} \cdot \mathbf{c})\mathbf{b}-(\mathbf{a} \cdot \mathbf{b})\mathbf{c},</math>
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| :<math>\left\|\mathbf{a}\times\mathbf{b}\right\|=\sqrt{\mathbf{a}^2\mathbf{b}^2-(\mathbf{a} \cdot \mathbf{b})^2}.</math>
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| ===Circumcenter coordinates===
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| ====Cartesian coordinates====
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| The [[Cartesian coordinates]] of the circumcenter are
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| :<math>U_x = ((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)) / D,</math>
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| :<math>U_y = ((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)) / D</math>
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| with | |
| :<math>D = 2( A_x(B_y - C_y) + B_x(C_y - A_y) + C_x(A_y - B_y)).\,</math>
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| 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 {{nowrap|''A''′ {{=}} ''A'' − ''A'' {{=}} (''A''′<sub>''x''</sub>,''A''′<sub>''y''</sub>) {{=}} (0,0)}}. In this case, the coordinates of the vertices {{nowrap|''B''′ {{=}} ''B'' − ''A''}} and {{nowrap|''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 coordinates of the triangle ''A''′''B''′''C''′ follow as
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| :<math>( C'_y(B^{'2}_x + B^{'2}_y) - B'_y(C^{'2}_x + C^{'2}_y) )/ D', \,</math>
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| :<math>( B'_x(C^{'2}_x + C^{'2}_y) - C'_x(B^{'2}_x + B^{'2}_y) )/ D' \,</math>
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| with
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| :<math>D' = 2( B'_xC'_y - B'_yC'_x ). \,</math>
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| ====Barycentric coordinates as a function of the side lengths====
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| The circumcenter has [[trilinear coordinates]] {{nowrap|(cos ''α'', cos ''β'', cos ''γ'')}} where {{nowrap|''α'', ''β'', ''γ''}} are the angles of the triangle. The circumcenter has [[barycentric coordinates (mathematics)|barycentric coordinates]]
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| :<math>\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), \,</math><ref>[http://mathworld.wolfram.com/BarycentricCoordinates.html Wolfram page on barycentric coordinates]</ref>
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| where {{nowrap|''a'', ''b'', ''c''}} are edge lengths ({{nowrap|''BC'', ''CA'', ''AB''}} respectively) of the triangle.
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| ===Barycentric coordinates from cross- and dot-products===
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| In [[Euclidean space]], there is a unique circle passing through any given three non-collinear points ''P''<sub>1</sub>, ''P''<sub>2</sub>, and ''P''<sub>3</sub>. Using [[Cartesian coordinates]] to represent these points as [[spatial vector]]s, it is possible to use the [[dot product]] and [[cross product]] to calculate the radius and center of the circle. Let
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| :<math>\mathrm{P_1} = \begin{bmatrix} x_1 \\ y_1 \\ z_1 \end{bmatrix},
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| \mathrm{P_2} = \begin{bmatrix} x_2 \\ y_2 \\ z_2 \end{bmatrix},
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| \mathrm{P_3} = \begin{bmatrix} x_3 \\ y_3 \\ z_3 \end{bmatrix}</math>
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| Then the radius of the circle is given by
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| :<math>\mathrm{r} = \frac
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| {\left|P_1-P_2\right| \left|P_2-P_3\right|\left|P_3-P_1\right|}
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| {2 \left|\left(P_1-P_2\right) \times \left(P_2-P_3\right)\right|}</math>
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| The center of the circle is given by the [[linear combination]]
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| :<math>\mathrm{P_c} = \alpha \, P_1 + \beta \, P_2 + \gamma \, P_3</math>
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| where
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| :<math>\alpha = \frac
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| {\left|P_2-P_3\right|^2 \left(P_1-P_2\right) \cdot \left(P_1-P_3\right)}
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| {2 \left|\left(P_1-P_2\right) \times \left(P_2-P_3\right)\right|^2}</math>
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| :<math>\beta = \frac
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| {\left|P_1-P_3\right|^2 \left(P_2-P_1\right) \cdot \left(P_2-P_3\right)}
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| {2 \left|\left(P_1-P_2\right) \times \left(P_2-P_3\right)\right|^2}</math>
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| :<math>\gamma = \frac
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| {\left|P_1-P_2\right|^2 \left(P_3-P_1\right) \cdot \left(P_3-P_2\right)}
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| {2 \left|\left(P_1-P_2\right) \times \left(P_2-P_3\right)\right|^2}</math>
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| ===Parametric equation of a triangle's circumcircle===
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| A [[unit vector]] [[perpendicular]] to the plane containing the circle is given by
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| :<math>\hat{n} = \frac
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| {\left( P_2 - P_1 \right) \times \left(P_3-P_1\right)}
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| {\left| \left( P_2 - P_1 \right) \times \left(P_3-P_1\right) \right|}.</math>
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| Hence, given the radius, ''r'', center, ''P<sub>c</sub>'', a point on the circle, ''P''<sub>0</sub> and a unit normal of the plane containing the circle, <math>\scriptstyle{\hat{n}}</math>, one parametric equation of the circle starting from the point ''P''<sub>0</sub> and proceeding in a positively oriented (i.e., [[right-hand rule|right-handed]]) sense about <math>\scriptstyle{\hat{n}}</math> is the following:
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| :<math>\mathrm{R} \left( s \right) = \mathrm{P_c} +
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| \cos \left( \frac{\mathrm{s}}{\mathrm{r}} \right) \left( P_0 - P_c \right) +
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| \sin \left( \frac{\mathrm{s}}{\mathrm{r}} \right)
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| \left[ \hat{n} \times \left( P_0 - P_c \right) \right].</math>
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| ===Angles===
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| {{anchor|Alternate segment theorem}}
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| {|align=center
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| |[[File:Circumcircle Angles 1.svg|160px|center]]
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| |width=50px|
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| |[[File:Circumcircle Angles 2.svg|160px|center]]
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| |}
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| 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 ''α'' meets the circle twice: once at each end; in each case at angle ''α'' (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.
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| ===Triangle centers on the circumcircle of triangle ABC===
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| In this section, the vertex angles are labeled ''A'', ''B'', ''C'' and all coordinates are [[trilinear coordinates]]:
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| *[[Steiner point]] = ''bc'' / (''b''<sup>2</sup> − ''c''<sup>2</sup>) : ''ca'' / (''c''<sup>2</sup> − ''a''<sup>2</sup>) : ''ab'' / (''a''<sup>2</sup> − ''b''<sup>2</sup>) = 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 {{nowrap|1/(''ax'') + 1/(''by'') + 1/(''cz'') {{=}} 0}}.)
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| *[[Tarry point]] = sec (''A'' + ω) : sec (''B'' + ω) : sec (''C'' + ω) = antipode of the Steiner point
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| *Focus of the [[Kiepert parabola]] = csc (''B'' − ''C'') : csc (''C'' − ''A'') : csc (''A'' − ''B'').
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| ===Other properties===
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| The circumcircle radius is no smaller than twice the incircle radius (Euler's triangle inequality).<ref name=Nelson>Nelson, Roger, "Euler's triangle inequality via proof without words," ''Mathematics Magazine'' 81(1), February 2008, 58-61.</ref>
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| The distance between the circumcenter and the incenter is <math>\sqrt{\scriptstyle{R(R-2r)}},</math> where ''r'' is the incircle radius and ''R'' is the circumcircle radius.<ref name=Nelson/>
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| The product of the incircle radius and the circumcircle radius of a triangle with sides ''a'', ''b'', and ''c'' is<ref>Johnson, Roger A., ''Advanced Euclidean Geometry'', Dover, 2007 (orig. 1929), p. 189, #298(d).</ref> <math>\tfrac{abc}{2(a+b+c)}.</math>
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| ==Cyclic quadrilaterals==
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| [[File:Cyclic quadrilateral.svg|thumb|right|300px|[[Cyclic quadrilateral]]s]]
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| {{main|Cyclic quadrilateral}}
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| Quadrilaterals that can be circumscribed have particular properties including the fact that opposite angles are [[supplementary angles]] (adding up to 180° or π radians).
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| ==Cyclic ''n''-gons==
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| For a cyclic polygon with an odd number of sides, all angles are equal if and only if the polygon is regular. A cyclic polygon with an even number of sides has all angles equal if and only if the alternate sides are equal (that is, sides 1, 3, 5, ... are equal, and sides 2, 4, 6, ... are equal).<ref>De Villiers, Michael. "Equiangular cyclic and equilateral circumscribed polygons," ''[[Mathematical Gazette]]'' 95, March 2011, 102-107.</ref>
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| A cyclic [[pentagon]] with [[rational number|rational]] sides and area is known as a [[Robbins pentagon]]; in all known cases, its diagonals also have rational lengths.<ref>{{citation
| |
| | last1 = Buchholz | first1 = Ralph H.
| |
| | last2 = MacDougall | first2 = James A.
| |
| | doi = 10.1016/j.jnt.2007.05.005
| |
| | issue = 1
| |
| | journal = [[Journal of Number Theory]]
| |
| | mr = 2382768
| |
| | pages = 17–48
| |
| | title = Cyclic polygons with rational sides and area
| |
| | url = http://docserver.carma.newcastle.edu.au/785/
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| | volume = 128
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| | year = 2008}}.</ref>
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| In any cyclic ''n''-gon with even ''n'', the sum of one set of alternate angles (the first, third, fifth, etc.) equals the sum of the other set of alternate angles. This can be proven by induction from the ''n''=4 case, in each case replacing a side with three more sides and noting that these three new sides together with the old side form a quadrilateral which itself has this property; the alternate angles of the latter quadrilateral represent the additions to the alternate angle sums of the previous ''n''-gon.
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| ==See also==
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| *[[Inscribed circle]]
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| *[[Jung's theorem]], an inequality relating the [[diameter]] of a point set to the radius of its minimum bounding circle
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| *[[Lester's theorem]]
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| *[[Circumscribed sphere]]
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| *[[Triangle center]]
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| *[[Japanese theorem for cyclic quadrilaterals]]
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| *[[Japanese theorem for cyclic polygons]]
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| ==Notes==
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| {{reflist|group="nb"}}
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| ==References==
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| <references/>
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| {{refbegin}}
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| *{{note|Coxeter}}{{cite book |author=Coxeter, H.S.M. |pages=12–13 |chapter=Chapter 1 |title=Introduction to geometry |publisher=Wiley |year=1969 |isbn=0-471-50458-0}}
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| *{{note|Megiddo}}{{cite journal
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| |author=Megiddo, N.
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| |title=Linear-time algorithms for linear programming in '''R'''<sup>3</sup> and related problems
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| |journal=SIAM Journal on Computing
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| |volume=12
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| |issue=4
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| |pages=759–776
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| |year=1983
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| |doi=10.1137/0212052}}
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| *{{cite journal
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| |author=Kimberling, Clark
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| |title=Triangle centers and central triangles
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| |journal=Congressus Numerantium
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| |volume=129
| |
| |year=1998
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| |pages=i–xxv, 1–295}}
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| *{{note|Pedoe}}{{cite book |author=[[Daniel Pedoe|Pedoe, Dan]] |title=Geometry: a comprehensive course |publisher=Dover |year=1988}}
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| {{refend}}
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| ==External links==
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| *[http://www.mathalino.com/reviewer/derivation-of-formulas/derivation-of-formula-for-radius-of-circumcircle Derivation of formula for radius of circumcircle of triangle] at Mathalino.com
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| * [http://dynamicmathematicslearning.com/semi-regular-anglegon.html Semi-regular angle-gons and side-gons: respective generalizations of rectangles and rhombi] at [http://dynamicmathematicslearning.com/JavaGSPLinks.htm Dynamic Geometry Sketches], interactive dynamic geometry sketch.
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| ===MathWorld===
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| *{{MathWorld |title=Circumcircle |urlname=Circumcircle}}
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| *{{MathWorld |title=Cyclic Polygon |urlname=CyclicPolygon}}
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| *{{MathWorld |title=Steiner circumellipse |urlname=SteinerCircumellipse}}
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| ===Interactive===
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| *[http://www.mathopenref.com/trianglecircumcircle.html Triangle circumcircle] and [http://www.mathopenref.com/trianglecircumcenter.html circumcenter] With interactive animation
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| *[http://www.uff.br/trianglecenters/X0003.html An interactive Java applet for the circumcenter]
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| [[Category:Circles]]
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| [[Category:Triangles]]
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| [[Category:Compass and straightedge constructions]]
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