Circle Defined by 3 Points Calculator
▲
X
_{1}
Y
_{1}
X
_{2}
Y
_{2}
X
_{3}
Y
_{3}
Center point (x,y):
Radius:
Area of circle:
Perimeter of circle:
Circle equation:
r
^{2}
Equation of a circle passing through
3
points
(
x
_{1}
, y
_{1}
) (
x
_{2}
, y
_{2}
)
and
(
x
_{3}
, y
_{3}
)
▲
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 coefficienta 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
(
x
_{1}
, y
_{1}
) (
x
_{2}
, y
_{2}
)
and
(
x
_{3}
, y
_{3}
)
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
)
.
The equations of two arbitrary lines are: y = m
_{1}
x + a and y = m
_{2}
x + b
Example:
Find the equation of a circle passing through the points
(− 3, 4) , (4, 5)
and
(1, − 4)
.
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 obtaine:
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: Ax
^{2}
+ Ay
^{2}
+ 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
(
x
^{2}
+
y
^{2}
) +
Bx
+
Cy
+
D
= 0
A
(
x
_{1}
^{2}
+
y
_{1}
^{2}
) +
Bx
_{1}
+
Cy
_{1}
+
D
= 0
A
(
x
_{2}
^{2}
+
y
_{2}
^{2}
) +
Bx
_{2}
+
Cy
_{2}
+
D
= 0
A
(
x
_{3}
^{2}
+
y
_{3}
^{2}
) +
Bx
_{3}
+
Cy
_{3}
+
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;
(
x
^{2}
+
y
^{2}
)
A
+
xB
+
yC
+
D
= 0
.
And the equation of the circle is:
x
^{2}
+
y
^{2}
−
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}
= a
^{2}
+ 2ab + b
^{2}
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}
=
5
^{2}