Let a circle be passing through the following three points.
(x_{1}, y_{1}), (x_{2}, y_{2}), (x_{3}, y_{3})
To get the equation of a circle passing through the three points, substitute the points into the general equation of a circle which is
x^{2} + x^{2} + 2gx + 2fy + c = 0
When you substitute the points (x_{1}, y_{1}), (x_{2}, y_{2}), (x_{3}, y_{3}), one by one into the above general equation of a circle, you will be getting three linear equations with three unknowns, which are g, f and c.
Solve the system of three linear equations for g, f and c.
Substitute the values of g, f and c into the general equation of the circle.
Example 1 :
Find the general equation the circle passing through the points (0, 1), (2, 3) and (-2, 5).
Solution :
The general equation of a circle :
x^{2} + y^{2} + 2gx + 2fy + c = 0
Substitute (0, 1).
0^{2} + 1^{2} + 2g(0) + 2f(1) + c = 0
1 + 2f + c = 0
2f + c = -1 ----(1)
Substitute (2, 3).
2^{2} + 3^{2} + 2g(2) + 2f(3) + c = 0
4 + 9 + 4g + 6f + c = 0
13 + 4g + 6f + c = 0
4g + 6f + c = -13 ----(2)
Substitute (-2, 5).
(-2)^{2} + 5^{2} + 2g(-2) + 2f(5) + c = 0
4 + 25 - 4g + 10f + c = 0
29 - 4g + 10f + c = 0
-4g + 10f + c = -29 ----(3)
(2) + (3) :
(4g + 6f + c) + (-4g + 10f + c) = (-13) + (-29)
4g + 6f + c - 4g + 10f + c = -13- 29
16f + 2c = -42
8f + c = -21 ----(4)
(1) - (4) :
(2f + c) - (8f + c) = (-1) - (-21)
2f + c - 8f - c = -1 + 21
-6f = 20
f = -¹⁰⁄₃
Substitute f = -¹⁰⁄₃ into (1).
2(-¹⁰⁄₃) + c = -1
-²⁰⁄₃ + c = -1
c = ¹⁷⁄₃
Substitute f = -¹⁰⁄₃ and c = ¹⁷⁄₃ nto (2).
4g + 6(-¹⁰⁄₃) + ¹⁷⁄₃ = -13
4g - 20 + ¹⁷⁄₃ = -13
4g = 7 - ¹⁷⁄₃
4g = ⁴⁄₃
g = ¹⁄₃
Substitute g = ¹⁄₃, f = -¹⁰⁄₃ and c = ¹⁷⁄₃ into the general equation of the circle.
x^{2} + y^{2} + 2(¹⁄₃)x + 2(-¹⁰⁄₃)y + ¹⁷⁄₃ = 0
x^{2} + y^{2} + (⅔)x - (²⁰⁄₃)y + ¹⁷⁄₃ = 0
Multiply both sides by 3.
3(x^{2} + y^{2} + (⅔)x - (²⁰⁄₃)y + ¹⁷⁄₃) = 3(0)
3x^{2} + 3y^{2} + 2x - 20y + 17 = 0
Example 2 :
Find the general equation the circle passing through the points (0, 1), (2, 3) and having centre on the line x - 2y + 3 = 0.
Solution :
The general equation of a circle :
x^{2} + y^{2} + 2gx + 2fy + c = 0
Substitute (0, 1).
0^{2} + 1^{2} + 2g(0) + 2f(1) + c = 0
1 + 2f + c = 0
2f + c = -1 ----(1)
Substitute (2, 3).
2^{2} + 3^{2} + 2g(2) + 2f(3) + c = 0
4 + 9 + 4g + 6f + c = 0
13 + 4g + 6f + c = 0
4g + 6f + c = -13 ----(2)
For the general equation of a circle
x^{2} + y^{2} + 2gx + 2fy + c = 0,
the center is given by
(-g, -f)
Since the circle has centre on the line x - 2y + 3 = 0, we can substitute (-g, -f) into the equation of the line.
-g - 2(-f) + 3 = 0
-g + 2f +3 = 0
-g + 2f = -3 ----(3)
(2) - (1) :
(4g + 6f + c - (2f + c) = (-13) - (-1)
4g + 6f + c - 2f - c = -13 + 1
4g + 4f = -12
g + f = -3 ----(4)
(3) + (4) :
(-g + 2f) + (g + f) = (-3) + (-3)
-g + 2f + g + f = -3- 3
3f = -6
f = -2
Substitute f = -2 into (4).
g - 2 = -3
g = -1
Substitute f = -2 into (1).
2(-2) + c = -1
-4 + c = -1
c = 3
Substitute g = -1, f = -2 and c = 3 into the general equation of the circle.
x^{2} + y^{2} + 2(-1)x + 2(-2)y + 3 = 0
x^{2} + y^{2} - 2x - 4y + 3 = 0
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