FIND THE RADIUS OF THE CIRCLE PASSES THROUGH THREE POINTS

Problem 1 :

Find the coordinates of the center of a circle passing through the points A(0, 0), B(-2, 1) and C(-3, 2). Also find the radius of this circle.

Solution:

The general equation of a circle 

x² + y² + 2gx + 2fy + c = 0

Equation of a circle which is passing through the point (0, 0).

0 + 0 + 2g(0) + 2f(0) + c = 0

c = 0

Equation of a circle which is passing through the point (-2, 1).

(-2)² + (1)² + 2g(-2) + 2f(1) + c = 0

5 - 4g + 2f + c = 0

-4g + 2f + c = -5 ----> (1)

Equation of a circle which is passing through the point (-3, 2).

(-3)² + (2)² + 2g(-3) + 2f(2) + c = 0

13 - 6g + 4f + c = 0

-6g + 4f + c = -13 ----> (2)

By applying c = 0 in (1) and (2)

-4g + 2f = -5 ---> (3)

-6g + 4f = -13 ---> (4)

(1) × 2 ==> -8g + 4f = -10 ----> (5)

Subtract (4) from this equation

-8g + 6g + 4f - 4f = -10 + 13

-2g = 3

g = -3/2

Substitute g = -3/2 in equation (3)

-4(-3/2) + 2f = -5

6 + 2f = -5

2f = -11

f = -11/2

Substitute g = -3/2, f = -11/2 and c = 0 in the general equation

x² + y² + 2gx + 2fy + c = 0

x² + y² + 2(-3/2)x + 2(-11/2)y + 0 = 0

x² + y² - 3x - 11y = 0

So, the equation of a circle passing through three points 

x² + y² - 3x - 11y = 0

Problem 2 :

Find the center of a circle passing through the points (6, -6), (3, -7) and (3, 3).

Solution:

The general equation of a circle

x² + y² + 2gx + 2fy + c = 0

Equation of a circle which is passing through the point (6, -6).

(6)² + (-6)² + 2g(6) + 2f(-6) + c = 0

36 + 36 + 12g - 12f + c = 0

72 + 12g - 12f + c = 0

12g - 12f + c = -72 ---> (1)

Equation of a circle which is passing through the point (3, -7).

(3)² + (-7)² + 2g(3) + 2f(-7) + c = 0

9 + 49 + 6g - 14f + c = 0

58 + 6g - 14f + c = 0

6g - 14f + c = -58 ---> (2)

Equation of a circle which is passing through the point (3, 3).

(3)² + (3)² + 2g(3) + 2f(3) + c = 0

9 + 9 + 6g + 6f + c = 0

18 + 6g + 6f + c = 0

6g + 6f + c = -18 ---> (3)

So, the three equations are

12g - 12f + c = -72 ---> (1)

6g - 14f + c = -58 ---> (2)

6g + 6f + c = -18 ---> (3)

By solving these three equations we can get the values of f, g and c.

By adding (1) and (2),

6g + 2f = -14 ---> (4)

By subtracting (2) and (3)

(6g - 14f + c) - (6g + 6f + c) = -58 + 18

-20f = -40

f = 2

Substitute f = 2 in equation (4),

6g + 2(2) = -14

6g + 4 = -14

6g = -18

g = -3

Substitute f = 2 and g = -3 in equation (1)

12g - 12f + c = -72

12(-3) - 12(2) + c = -72

-36 - 24 + c = -72

-60 + c = -72

c = -72 + 60

c = -12

Substitute g = -3, f = 2 and c = -12 in the general equation

x² + y² + 2gx + 2fy + c = 0

x² + y² + 2(-3)x + 2(2)y - 12 = 0

x² + y² - 6x + 4y - 12= 0

So, the equation of a circle passing through three points 

x² + y² - 6x + 4y - 12 = 0

Problem 3 :

Find the coordinates of the point equidistant from three given points A(5, 3), B(5, -5) and C(1, -5).

Solution:

The general equation of a circle

x² + y² + 2gx + 2fy + c = 0

Equation of a circle which is passing through the point (5, 3).

(5)² + (3)² + 2g(5) + 2f(3) + c = 0

25 + 9 + 10g + 6f + c = 0

34 + 10g + 6f + c = 0

10g + 6f + c = -34 ---> (1)

Equation of a circle which is passing through the point (5, -5).

(5)² + (-5)² + 2g(5) + 2f(-5) + c = 0

25 + 25 + 10g - 10f + c = 0

50 + 10g - 10f + c = 0

10g - 10f + c = -50 ---> (2)

Equation of a circle which is passing through the point (1, -5).

(1)² + (-5)² + 2g(1) + 2f(-5) + c = 0

1 + 25 + 2g - 10f + c = 0

26 + 2g - 10f + c = 0

2g - 10f + c = -26 ---> (3)

So, the three equations are

10g + 6f + c = -34 ---> (1)

10g - 10f + c = -50 ---> (2)

2g - 10f + c = -26 ---> (3)

By subtracting (1) and (2)

(10g + 6f + c) - (10g - 10f + c) = -34 + 50

16f = 16

f = 1

By subtracting (2) and (3)

(10g - 10f + c) - (2g - 10f + c) = -50 + 26

8g = -24

g = -3

Substitute f = 1 and g = -3 in equation (1)

10(-3) + 6(1) + c = -34

-30 + 6 + c = -34

-24 + c = -34

c = -10

Substitute g = -3, f = 1 and c = -10 in the general equation

x² + y² + 2gx + 2fy + c = 0

x² + y² + 2(-3)x + 2(1)y - 10 = 0

x² + y² - 6x + 2y - 10 = 0

So, the equation of a circle passing through three points 

x² + y² - 6x + 2y - 10 = 0.