On this page "Equation of the circle" we are going to see the methods to find the equation of any circle.

Actually we have three methods to find equation of any circle.

(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

(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 a 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).

**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

**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.

**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.

**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

x² + y² + (2/3)x + (-20/3)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

**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=5****²**

**(x+2)²+y²=25**

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

**(x-6)²+(y-0)²=5****²**

**(x-6)²+y²=25**

Related Topics

- With center and radius
- With endpoints of a diameter
- Equation of a circle passing though three points
- Length of the tangent to a circle
- Equation of the tangent to a circle
- Family of circles
- Orthogonal circles
- Section formula
- Area of triangle
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- Finding missing vertex using centroid worksheet
- Midpoint
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- Orthocentre of a triangle
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- Locus
- Perpendicular distance
- Angle between two straight lines
- Parabola
- Ellipse

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