EQUATION OF A CIRCLE PASSING THROUGH THREE POINTS

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 = ¹⁄₃, f = -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 + (⅔)x - (²⁰⁄₃)y + ¹⁷⁄₃ = 0

Multiply both sides by 3.

3(x² + y² + (⅔)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

Example 3 :

The circle C has equation

x² + y² - 6x + 4y = 12

a) Find the center and radius of C.

b) The point P(-1, 1) and the point Q(7, -5) both lie on C.

(b) Show that PQ is a diameter of C. 

Solution :

a)

x² + y² - 6x + 4y = 12

x² + y² - 6x + 4y - 12 = 0

x² - 6x + y² + 4y - 12 = 0

(x - 3)² - 3²+ (y + 2)² - 2² - 12 = 0

(x - 3)² - 9 + (y + 2)² - 4 - 12 = 0

(x - 3)² + (y + 2)² - 25 = 0

(x - 3)² + (y + 2)² = 25

Center of the circle is (3, -2).

b) The point P(-1, 1) and the point Q(7, -5) both lie on C.

c) Equation of line joining the points P(-1, 1) and Q(7, -5)

Slope = (-5 - 1)/(7 + 1)

= -6/8

= -3/4

(y - y₁) = m(x - x₁)

(y - 1) = -3/4(x + 1)

4(y - 1) = -3(x + 1)

4y - 4 = -3x - 3

4y = -3x - 3 + 4

4y = -3x + 1

y = (-3/4)x + (1/4)

Applying the center point (3, -2), we get

-2 = (-3/4) (3) + (1/4)

-2 = -9/4 + 1/4

-2 = -8/4

-2 = -2

So, the center point lies on the line and it is the diameter of the circle.

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