We have three methods to find equation of any circle.

**Method 1 :**

(1) Equation of circle when **center is at origin and radius (r) is given **

x² + y² = r²

here "r" represents the radius of the circle

**Method 2 :**

(2) Equation of circle when **center is at (h,k) and radius (r) is given **

(x - h)² + (y - k)² = r²

here (h,k) represents center of the circle and "r" represents radius of the circle.

(3) Equation of circle when **the endpoints of the diameter (x****₁ ,y₁) and (x₂,y₂) are given **

(x-x₁)(x-x₂)+(y-y₁)(y-y₂) = 0

here (x₁ ,y₁) and (x₂,y₂) represents the endpoints of the diameter of the circle.

**Problem 1 :**

Find the equation of the circle if the center is (2,-3) and radius is 4 units**Solution :**

Equation of circle with center (h,k) and radius "r" is given

**(x-h)²+(y-k)²=r²**

h = 2,k = -3 and radius (r) = 4 units

(x-2)²+(y-(-3))²= 4²

(x-2)²+(y+3)²= 4²

expanding (x-2)² and (y+3)² using the __algebraic identities__**(a-b)****²** and **(a+b)**** ² **we get,

x² + 2² - 2(x)(2) + y² + 3² + 2(y)(3) = 16

x² + 4 - 4x + y² + 9 + 6y = 16

x² - 4x + y² + 6y + 13 - 16 = 0

x² - 4x + y² + 6y - 3 = 0

**Problem 2 :**

Find the equation of the circle if the center is (1,5) and radius 9 and check whether the circle passes through the point (2,0)**Solution :**

The equation of the circle is (x-h)² + (y-k)² = r²

(h,k) = (1,5) and r = 9

**Equation of circle :**

(x-h)² + (y-k)² = r²

(x-1)² + (y-5)² = 9²

x² + 1² - 2(x)(1) + y² + 5² + 2(y)(5) = 81

x² + 1 - 2x + y² + 25 + 10y = 81

x² - 2x + y² + 25y + 26 - 81 = 0

x² - 2x + y² + 25y -55 = 0

To check whether the circle passes through the point (2,0), we have to apply the values x = 2 and y = 0 in the equation of a circle.

2² - 2(2) + 0² + 25(0) - 55 = 0

4 - 4 -55 = 0

-55 ≠ 0

The equation is not satisfying the condition. So the circle is not passing through the point (2,0).

Let us look into the next example problem "How to find equation of a circle".

**Problem 3 :**

Find the equation of the circle if the center is (1,-3) and passing through the point (4,1)

**Solution :**

**The equation of the circle is (x-h)²+(y-k)² = r²**

Here we have only the center of the circle.We don't have the radius.To find that let us make diagram with the given details.

To find the radius of a circle we have to find the distance from O to A.

Distance of OA (or) radius of a circle

OA = √(x₂ - x₁)² + (y₂ - y₁)²

Here x₁ = 1 x₂ = 4 y₁ = -3 y₂ = 1

OA = √(4 - 1)² + (1 - (-3))²

OA = √(3)² + (1 +3)²

OA = √9 + 4²

OA = √9 + 16

OA = √25

OA = √5 x 5

OA = 5 units

Now we have center and radius

**Equation of the circle :**

(x-h)² + (y-k)² = r²

(x-1)² + (y-(-3))² = 5²

(x-1)² + (y+3)² = 25

x² + 1² - 2(x)(1) + y² + 3² + 2(y)(3) = 25

x² + 1 - 2x + y² + 9 + 6y = 25

x² - 2x + y² + 6y + 10 - 25 = 0

x² - 2x + y² + 6y - 15 = 0

Let us look into the next example problem "How to find equation of a circle".

**Problem 4 :**

Find the equation of the circle if (2,-1) and (5,-3) are the endpoints of the diameter.

**Solution :**

Now we have to consider the given points as (x₁,y₁) and (x₂,y₂). So the values of x₁ = 2,y₁ = -1,x₂ = 5 and y₂ = -3

**Equation of the circle :**

**(x-x₁) (x-x₂) + (y-y₁) (y-y₂) = 0**

( x - 2) (x - 5) + (y - (-1)) (y - (-3)) = 0

( x - 2) (x - 5) + (y + 1)) (y + 3)) = 0

x² - 2x - 5x + 10 + y² +y +3y + 3 = 0

x² - 7x + 10 + y² + 4y + 3 = 0

x² - 7x + y² + 4y + 10 + 3 = 0

x² - 7x + y² + 4y + 13 = 0

So the required equation of the circle x²-7x + y²+4y+13=0.

Let us look into the next example problem "How to find equation of a circle".

**Problem 5 :**

Find the equation of a circle if (1,3) and (2,0) are the endpoints of the diameter.

**Solution :**

Now we have to consider the given points as (x₁,y₁) and (x₂,y₂). So the values of x₁ = 1,y₁ = 3,x₂ = 2 and y₂ = 0

Formula to find the equation of circle if the endpoints of a diameter are given.

**(x-x₁) (x-x₂) + (y-y₁) (y-y₂) = 0**

( x - 1) (x - 2) + (y - 3) (y - 0) = 0

( x - 1) (x - 2) + (y - 3) y = 0

x² - 2x - x + 2 + y² - 3y = 0

x² - 3x + y² - 3y + 2 = 0

So the required equation of the circle x²-3x+y²-3y+2=0.

Let us look into the next example problem "How to find equation of a circle".

**Problem 6:**

Find the equation of a 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**

Now we have to apply these points one by one in the above equation of circle

equation of a circle which is passing through the point (0,1)

0² + 1² + 2g(0) + 2 f(1) + c = 0

1 + 0 + 2f + c = 0

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

equation of a circle which is passing through the point (2,3)

2² + 3² + 2g(2) + 2 f(3) + c = 0

4 + 9 + 4g + 6f + c = 0

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

equation of a circle which is passing through the point (-2,5)

(-2)² + 5² + 2g(-2) + 2 f(5) + c = 0

4 + 25 - 4g + 10f + c = 0

-4g + 10f + c = -29 ----------(3)

So the three equations are

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

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

-4g + 10f + c = -29 ----------(3)

by solving these three equations we can get the values of f,g and c by adding (2) and (3)

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

-4g + 10f + c = -29 ----------(3)

----------------------

16f + 2c = -42 ---------(5)

Multiplying (1) by 2

4f + 2c = -2

Subtract (5) form this equation

16f + 2c = -42

4f + 2c = -2

(-) (-) (+)

----------------

12f = -40

f = -40/12

f = -10/3

Substitute f = -10/3 in the first equation

2(-10/3) + c = -1

-20/3 + c = -1

c = -1 + 20/3

c = 17/3

substitute c = 17/3 and f = -10/3 in the second equation

4g + 6(-10/3) + 17/3 = -13

4g - 20 + 17/3 = - 13

4g + (-60+17)/3 = -13

4g - 43/3 = -13

4g = -13 + 43/3

4g = (-39+43)/3

4g = 4/3

g = 4/(4x3)

g = 1/3

Substitute g = 1/3 c = 17/3 ad f = -10/3 in the general equation

**x² + y² + 2gx + 2fy + c = 0**

x² + y² + 2(1/3)x + 2(-10/3)y + 17/3 = 0

3x² + 3y² + 2x -20y + 17 = 0

So the equation of a circle passing through three points 3x² + 3y² + 2x -20y + 17 = 0

Let us look into the next example problem "How to find equation of a circle".

**Problem 7 :**

Find the equation of a circle which passes through (2,3) and whose center is on x axis and radius is 5 units.

**Solution :**

Equation of a circle

**(x-h)²+(y-k)²=r****²**

equation of the circle passing through the point (2,3) is

(2 -h)²+(3-k)² = 5²

since the center point is on x axis,the y- coordinate value will be 0.So the center is **(h,0)**

(2 -h)²+(3-0)² = 5²

(2 -h)²+ 9 = 25

(2 -h)² = 16

2 - h = ± 4

2 - h = 4 2 - h = -4

- h = 4 - 2 - h = - 4 - 2

- h = 2 - h = - 6

h = -2 and h = 6

Equation of circle with center (-2,0) and radius 5 units

(x+2)²+(y-0)²=5²

(x+2)²+y²=25

Equation of circle with center (6,0) and radius 5 units

(x-6)²+(y-0)²=5²

(x-6)²+y²=25

Related Topics

- With center and radius
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- Locus
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- Angle between two straight lines
- Parabola
- Ellipse

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