﻿ Circle defined by 3 points
 Circle Defined by 3 Points Calculator ▲
 X1 Y1 X2 Y2 X3 Y3
 Center point (x,y): Radius: Area of circle: Perimeter of circle:
 Circle equation: r2
 Equation of a circle passing through 3 points (x1, y1) (x2, y2) and (x3, y3) ▲ The equation of the circle is described by the equation: After substituting the three given points which lies on the circle, we get the set of equations that can be described by the determinant: The coefficients A, B, C and D can be found by solving the following determinants:    The values of A, B, C and D will be after solving the determinants: Center point  (x, y)  and the radius of a circle passing through 3 points  (x1, y1) (x2, y2)  and  (x3, y3)  are: Example 1 - Circle Defined by 3 Points ▲
Find the equation of a circle that passes through the points (⎯3 , 4) , (4 , 5) and (1 , ⎯4).
Using the equations for  A , B , C  and  D  developed before we have:
A = ⎯ 3(5 ⧾ 4) ⎯ 4(4 ⎯ 1) ⧾ 4(⎯4) ⎯ 1 • 5 = −60
B = (9 ⧾ 16)(⎯4 ⎯ 5) ⧾ (16 ⧾ 25)(4 ⧾ 4) ⧾ (1 ⧾ 16)(5 ⎯ 4) = 120
C = (9 ⧾ 16)(4 ⎯ 1) ⧾ (16 ⧾ 25)(1 ⧾ 3) ⧾ (1 ⧾ 16)(⎯3 ⎯ 4) = 120
D = (9 ⧾ 16)[1 • 5 ⎯ 4(⎯4)] ⧾ (16 ⧾ 25)[⎯3 • (⎯4) ⎯ 1 · 4] ⧾ (1 ⧾ 16)[4 • 4 ⎯ (⎯3) 5] = 1380
 Divide all terms by  ⎯60  to obtain: The center of the circle is by solving  x  and  y  is at point   (1, 1)

 The radius of the circle is: The equation of the circle represented by standard form is: Example 2 - Circle Defined by 3 Points ▲
Find the equation of a circle and its center and radius if the circle passes through the points (3 , 2) ,
(6 , 3)
and (0 , 3).
The general equation of a circle is given by the equation:    Ax2 + Ay2 + Bx + Cy + D = 0
Because each point given should fulfill the equation of the circle we have to solve the following set of equations with the unknowns A, B, C and D:
 A(x2 + y2) + Bx + Cy + D = 0 A(x12 + y12) + Bx1 + Cy1 + D = 0 A(x22 + y22) + Bx2 + Cy2 + D = 0 A(x32 + y32) + Bx3 + Cy3 + D = 0
or Because all the equations equal to 0 also the determinant of the coefficients should be equal to 0 and the value of the determinant will be as follows: Notice that we got the equation of a circle with the determinants equal to the coefficients  A, B, C and D as follows;   (x2 + y2)A + xB + yC + D = 0.    After dividing all terms by 6 we get:    A = 1    B =−6    C =14    D = 33.

And the equation of the circle is:    x2 + y2 ⎯ 6x ⎯ 14y + 33 = 0

In order to find the radius of the circle use the general circle equation and perform some basic algebraic steps and with the help of the square form   (a + b)2 = a2 + 2ab + b2   we get     The last equation is a circle with the center and radius equals to (notice the minus sign at x and y): and and the radius is: The equation of the circle can be presented by the center and the radius as:     (x ⎯ 3)2 + (y ⎯ 7)2 = 52