The circle which is drawn around a polygon touching all the vertices of the polygon is called the Circumcircle. It is also called Circumscribed circle

Circumcircle is the smallest circle passing through each of the corner points of a regular polygon. The radius of the circumcircle is equal to the radius of the regular polygon. It is called the Circumradius. 

The center of the Circumcircle is the intersection of the diagonals of the regular polygon called Circumcenter. It coincides with the center of the regular polygon

Circumcircle

$OB$ = $Circumradius,\ O$ = $Circumcenter$
The formulas are of the following:
• Radius of the Circumcircle

$Radius$ = $\frac{m}{2 \times Sin (\frac{p}{n})}$

$m$ = $length\ of\ any\ side\ of\ regular\ polygon$

$n$ = $number\ of\ sides\ of\ regular\ polygon$
• Area of Circumcircle

$Area\ =\ p\ \times\ r^{2}$

$r$ = $radius\ of\ circumcircle$
• Circumcenter - Circumcenter of triangle ABC is the intersection point of the perpendicular bisectors of the sides of a triangle. It lies at an equal distance from each of the three vertices. 

Depending on the type of triangle, the circumcenter lies as below:

i) In an acute angled triangle the circumcenter lies inside the triangle

ii) In an obtuse angled triangle the circumcenter lies outside the triangle

iii) In a right angled triangle the circumcenter lies at the hypotenuse of the triangle

• Radius - Radius of Circumcircle can be expressed as follows:

$r$ = $\frac{pqr}{\sqrt{(p^{2} + q^{2} + r^{2}) -2 (p^{4} + q^{4} + r^{4}})}$

$r$ = $\frac{pqr}{4A}$

Where, A is the area of the triangle

$r$ = $\frac{p}{2SinP}$ = $\frac{q}{2SinQ}$ = $\frac{r}{2SinR}$

Where, p, q and r are the sides of the triangle

P, Q and R are the three angles of the triangle 
• Diameter - Diameter of Circumcircle satisfies the Sine Rule

$D$ = $\frac{p}{SinP}$ = $\frac{q}{SinQ}$ = $\frac{r}{SinR}$

$D$ = $\frac{pqr}{2A_{T}}$

$A_{T}$ = $Area\ of\ triangle$ = $\sqrt{s(s-p)(s-q)(s-r)}$

$s$ = $Semi\ perimeter$ = $\frac{(p+q+r)}{2}$
• Area - Area of Circumcircle can be expressed as follows:

$A$ = $\frac{p^{2} q^{2} r^{2}}{(p^{2} + q^{2} + r^{2})\ -2\ (p^{4} + q^{4} + r^{4})}$

$A_{C}$ = $\frac{p^{2} q^{2} r^{2} \pi}{16A_{T}^{2}}$

Where, $A_{T}$ = $\frac{pqSinR}{2}$ = $\frac{qrSinP}{2}$ = $\frac{rpSinQ}{2}$
The smallest circle possible through the four vertices of the square is called the Circumcircle of a Square. The center of the Circumcircle is the intersection point of the two diagonals of the square, called Circumcenter of a Square.

Circumcircle of a Square

$AB$ = $BC$ = $CD$ = $DA$ = Side of Square

$O$ = Circumcenter

$AC$ = $BD$ = Diagonal of Square

Radius of Circumcircle of Square

$r$ = $\frac{a}{\sqrt{2}}$

$a$ = Side of square

$r$ = $\frac{D}{2}$

$D$ = Diagonal of square

Area of Circumcircle of Square

$A_{c}$ = $\pi r^{2}$

$r$ = Radius of Circumcircle

$A_{C}$ = $\frac{\pi A_{s}}{2}$

$A_{C}$ = Area of Circumcircle

$A_{S}$ = Area of square
The smallest circle possible through the three vertices of a triangle whose all three sides are equal is called Circumcircle of an Equilateral Triangle. 

The center of the Circumcircle is the intersection of the medians of the equilateral triangle, called Circumcircle of Equilateral Triangle.

Circumcircle of Equilateral Triangle

$AB$ = $BC$ = $CA$ = Side of Equilateral triangle

$O$ = Circumcenter

• Radius of Circumcircle of Equilateral Triangle

$r$ = $\frac{a}{\pi 3}$

$a$ = Side of Equilateral Triangle

$r$ = $2\ \times$ $\frac{m}{3}$

$m$ = Median of Equilateral triangle

• Area of Circumcircle of Equilateral Triangle

$A_{C}$ = $\pi r^{2}$

$r$ = Radius of Circumcircle 

$A_{C}$ = $\frac{4 \pi A_{T}}{3 \sqrt{3}}$

$A_{C}$ = Area of Circumcircle

$A_{T}$ = Area of triangle

Circumcircle Equation

• Circumcircle Equation formed by Pedoe in the year 1995 can be represented as follows:

$\begin{vmatrix} x^{2} + y^{2} & x & y & 1\\ x_{1}^{2} + y_{1}^{2} & x_{1} & y_{1} & 1\\ x_{2}^{2} + y_{2}^{2} & x_{2} & y_{2} & 1\\ x_{3}^{2} + y_{3}^{2} & x_{3} & y_{3} & 1 \end{vmatrix} = 0$

• Circumcenter

$x_{0}$ = $\frac{1}{2p}$ $\begin{vmatrix} x_{1}^{2} + y_{1}^{2} & y_{1} & 1 \\ x_{2}^{2} + y_{2}^{2} & y_{2} & 1 \\ x_{3}^{2} + y_{3}^{2} & y_{3} & 1 \end{vmatrix}$

$y_{0}$ = $\frac{-1}{2p}$ $\begin{vmatrix}x_{1}^{2} + y_{1}^{2} & x_{1} & 1\\ x_{2}^{2} + y_{2}^{2} & x_{2} & 1\\ x_{3}^{2} + y_{3}^{2} & x_{3} & 1 \end{vmatrix}$

• Circumradius

$r$ = $\frac{\sqrt{q_{x}^{2} + q_{y}^{2} - 4qr}}{2|p|}$

$Where,\ q_{x}$ = $\begin{vmatrix}x_{1}^{2} + y_{1}^{2} & y_{1} & 1\\ x_{2}^{2} + y_{2}^{2} & y_{2} & 1\\ x_{3}^{2} + y_{3}^{2} & y_{3} & 1\end{vmatrix}$

$and\ q_{y}$ = $\begin{vmatrix}x_{1}^{2} + y_{1}^{2} & x_{1} & 1\\ x_{2}^{2} + y_{2}^{2} & x_{2} & 1\\ x_{3}^{2} + y_{3}^{2} & x_{3} & 1\end{vmatrix}$
Example 1:  

A hexagon has sides each of length 8 inches. Please find out the radius and area of its circumcircle

Solution:

$Radius$ =   $\frac{m}{2\ \times\ Sin (\frac{\pi}{n})}$

m = length of hexagon = 8 inches

n = number of sides =6

Therefore, Radius = $\frac{8}{2\ \times\ Sin (\frac{\pi}{6})}$

= $\frac{8}{2\ \times\ \frac{1}{2}}$

= 8 inches

Area = $\pi \times r^{2}$

= $\pi \times 8^{2}$

= $64\pi\ square\ inches$
Example 2:

A triangle has its sides 5 inches, 12 inches and 13 inches respectively. The corresponding angles are 30, 60 and 90 degrees. Please find out the radius, diameter and area of the circumcircle

Circumcircle Example

Solution

Given, p = 5, q = 12, r = 13, P = 30, Q = 60, R = 90

Radius:

$r$ = $\frac{q}{2SinQ}$

$r$ = $\frac{13}{2Sin30}$
   
$r$ = $\frac{13}{2 \times \frac{1}{2}}$

$r$ = $6.5\ inches$

Diameter:  
       
       $\frac{pqr}{2A_{T}}$

$A_{T}$ = $\frac{5 \times 12}{2}$ = 30 square inches

Diameter = $\frac{(5 \times 12 \times 13)}{(30)}$ = $13\ inches$

Area:

$A_{C}$ = $\frac{p^{2} q^{2} r^{2}\ \pi}{16A_{T}^{2}}$

   = $\frac{5^{2} \times 12^{2} \times 13^{2}}{16 \times 30^{2}}$

   = $\frac{25 \times 144 \times 169}{16 \times 900}$

   = $42.25\ square\ inches$
Example 3: 

A square has length 4 cm each. Find out its radius, diagonal and area of its circumcircle

Circumcircle Examples

Solution:  

Radius:

$r$ = $\frac{a}{\sqrt2}$, $a$ = Side of square

$r$ = $\frac{4}{\sqrt2}$

$r$ = $\frac{2}{\sqrt2}$cm

Diagonal: $d$ = $\sqrt{2a}$ = $\sqrt{2} \times 4$ = $4\sqrt{2}$cm

Area:

$A_{C}$ = $\frac{\pi A_{S}}{2}$

$A_{S}$ = $4^{2}$ = $16$ square centimeter

$A_{C}$ = $\frac{16 \pi}{2}$ = $8 \pi$ square centimeter.
Example 4:

An equilateral triangle has side of 3cm. Find out the radius and area of its circumcircle

Circumcircle Problem

Solution: 

Radius:

$r$ = $\frac{a}{\sqrt3}$

= $\frac{3}{\sqrt3}$cm

= $\sqrt3$cm

Area:
$A_{C}$ = $\frac{4 \pi A_{T}}{3 \sqrt3}$

$A_{T}$ = $\frac{\sqrt3 a^{2}}{4}$ 

      = $\frac{\sqrt3 \times 3^{2}}{4}$

      = $\frac{9 \sqrt3}{4}$ square centimeter