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

πr=  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

x+ y2  - 6x - 4y + 4  =  0

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