# EQUATION OF A CIRCLE PASSING THROUGH THREE POINTS

Let a circle be passing through the following three points.

(x1, y1), (x2, y2), (x3, y3)

To get the equation of a circle passing through the three points, substitute the points into the general equation of a circle which is

x2 + x2 + 2gx + 2fy + c = 0

When you substitute the points (x1, y1), (x2, y2), (x3, y3), 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 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 :

x2 + y2 + 2gx + 2fy + c = 0

Substitute (0, 1).

02 + 12 + 2g(0) + 2f(1) + c = 0

1 + 2f + c = 0

2f + c = -1 ----(1)

Substitute (2, 3).

22 + 32 + 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 + 52 + 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 = ¹⁄₃f = -¹⁰⁄₃ and c = ¹⁷⁄₃ into the general equation of the circle.

x2 + y2 + 2(¹⁄₃)x + 2(-¹⁰⁄₃)y + ¹⁷⁄₃ = 0

x2 + y2 + ()x - (²⁰⁄₃)y + ¹⁷⁄₃ = 0

Multiply both sides by 3.

3(x2 + y2 + ()x - (²⁰⁄₃)y + ¹⁷⁄₃) = 3(0)

3x2 + 3y2 + 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 :

x2 + y2 + 2gx + 2fy + c = 0

Substitute (0, 1).

02 + 12 + 2g(0) + 2f(1) + c = 0

1 + 2f + c = 0

2f + c = -1 ----(1)

Substitute (2, 3).

22 + 32 + 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

x2 + y2 + 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.

x2 + y2 + 2(-1)x + 2(-2)y + 3 = 0

x2 + y2 - 2x - 4y + 3 = 0

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