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Circle defined by 3 points calculator

Print circle 3 points
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) summary

Print 3 points circle summary
The equation of the circle is described by the equation:
Circle 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:
Circle equation

The coefficients A, B, C and D can be found by solving the following determinants:

Circle equation A
Circle equation B
Circle equation C
Circle equation D
Circle equation

Center point  (x, y)  and the radius of a circle passing through 3 points  (x1, y1) (x2, y2)  and  (x3, y3)  are:

Circle equation

Example 1 - Circle Defined by 3 Points

Ex-1
Example 1 - Circle Defined by 3 Points Print example of 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

Ex-2
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 (█(■(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 ))(■(A@B@■(C@D)))=0
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:
(x^2+y^2 )|■(x_1&y_1&1@x_2&y_2&1@x_3&y_3&1)|-x|■(〖x_1〗^2+〖y_1〗^2&y_1&1@〖x_2〗^2+〖y_2〗^2&y_2&1@〖x_3〗^2+〖y_1〗^2&y_3&1)|+y|■(〖x_1〗^2+〖y_1〗^2&x_1&1@〖x_2〗^2+〖y_2〗^2&x_2&1@〖x_3〗^2+〖y_1〗^2&x_3&1)|-|■(〖x_1〗^2+〖y_1〗^2&x_1&y_1@〖x_2〗^2+〖y_2〗^2&x_2&y_2@〖x_3〗^2+〖y_1〗^2&x_3&y_3 )|
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.
A=|■(3&2&1@6&3&1@0&3&1)|=6 B=|■(13&2&1@45&3&1@9&3&1)|=36 C=|■(13&3&1@45&6&1@9&0&1)|=-84 D=|■(13&3&2@45&6&3@9&0&3)|=-198
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
(Ax^2+Bx)+(Ay^2+Cy)+D=0
A(x^2+B/A x)+A(y^2+C/A y)+D=0
A[(x+B/2A)^2-B^2/(4A^2 )]+A[(y+C/2A)^2-C^2/(4A^2 )]+D=0
[(x+B/2A)^2-B^2/(4A^2 )]+[(y+C/2A)^2-C^2/(4A^2 )]=-D/A
(x+B/2A)^2+(y+C/2A)^2=B^2/(4A^2 )+C^2/(4A^2 )-D/A
The last equation is a circle with the center and radius equals to (notice the minus sign at x and y):
x=-B/2A and y=-C/2A
and the radius is: r=√(B^2/(4A^2 )+C^2/(4A^2 )-4D/4A)=√((B^2+C^2)/(4A^2 )-D/A)
The equation of the circle can be presented by the center and the radius as:     (x ⎯ 3)2 + (y ⎯ 7)2 = 52