**Equation of Circle Passes Through 3 Points :**

Here we are going to see, how to find equation of circle passes through three given points

**Question 1 :**

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

**Solution :**

Equation of circle

x^{2} + y^{2} + 2gx + 2fy + c = 0 ----(A)

The point (1, 0) lies on the circle

1^{2} + 0^{2} + 2g(1) + 2f(0) + c = 0

1 + 2g + c = 0

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

The point (-1, 0) lies on the circle

(-1)^{2} + 0^{2} + 2g(-1) + 2f(0) + c = 0

1 - 2g + c = 0

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

The point (0, 1) lies on the circle

0^{2} + 1^{2} + 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

x^{2} + y^{2} + 2(0)x + 2(0)y - 1 = 0

x^{2} + y^{2} = 1

**Question 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 = πr^{2}

πr^{2 }= 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} = r^{2}

(x - 3)^{2} + (y - 2)^{2} = 3^{2}

x^{2} - 6x + 9 + y^{2} - 4y + 4 - 9 = 0

x^{2 }+ y^{2} - 6x - 4y + 4 = 0

After having gone through the stuff given above, we hope that the students would have understood, "Equation of Circle Passes Through 3 Points".

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