EQUATION OF CIRCLE PASSES THROUGH 3 POINTS

Example 1 :

Find the equation of the circle through the points (1, 0), (-1, 0) and (0, 1) .

Solution :

Equation of circle

x² + y² + 2gx + 2fy + c = 0  ----(A)

The point (1, 0) lies on the circle 

1² + 0² + 2g(1) + 2f(0) + c = 0

1 + 2g + c  =  0

2g + c  =  -1  ----(1)

The point (-1, 0) lies on the circle 

(-1)² + 0² + 2g(-1) + 2f(0) + c = 0

1 - 2g + c  =  0

-2g + c  =  -1  ----(2)

The point (0, 1) lies on the circle 

0² + 1² + 2g(0) + 2f(1) + c = 0

1 + 2f + c  =  0

2f + c  =  -1  ----(3)

(1) + (2)

2c  =  -2

c  =  -1

By applying c = -1 in (1), we get 

2g -1  =  -1

2g  =  0

g  =  0

By applying c = -1 in (3), we get 

2f -1  =  -1

2f  =  0

f  =  0

By applying the values of f, g and c in (A), we get

x² + y² + 2(0)x + 2(0)y - 1 = 0

x² + y²  = 1

Example 2 :

A circle of area 9π  square units has two of its diameters along the lines x + y = 5 and x − y = 1. Find the equation of the circle.

Solution :

Area of circle  =  πr²

πr²  =  9π

r  =  3

By solving the equations of diameters of circle, we get the center.

 x + y = 5 ----(1)

 x − y = 1----(2)

(1) + (2)

2x  =  6  ==>  x  =  3

3 + y  =  5

y  =  5 - 3

y  =  2

Center of the circle (3, 2)

Equation of the circle 

(x - h)² + (y - k)²  =  r²

(x - 3)² + (y - 2)²  =  3²

x² - 6x + 9 + y² - 4y + 4 - 9  =  0

x² + y²  - 6x - 4y + 4  =  0

Kindly mail your feedback to v4formath@gmail.com

We always appreciate your feedback.