# Equation of a Circle

See how to write and use the equation of a circle.

6 examples and their solutions.

## Equation of a Circle

### Formula

(x - h)

^{2}+ (y - k)

^{2}= r

^{2}

### Example

Center: (2, 1), Radius: 3

Equation of the circle?

Solution Equation of the circle?

(2, 1)

r = 3

(x - 2)

(x - 2)

r = 3

(x - 2)

^{2}+ (y - 1)^{2}= 3^{2}(x - 2)

^{2}+ (y - 1)^{2}= 9Close

### Example

Center: (4, 0), Diameter: 14

Equation of the circle?

Solution Equation of the circle?

(4, 0)

d = 14

→ r = 7

(x - 4)

(x - 4)

d = 14

→ r = 7

(x - 4)

^{2}+ (y - 0)^{2}= 7^{2}(x - 4)

^{2}+ y^{2}= 49Close

### Example

Endpoints of the diameter: (-1, 3), (7, 1)

Equation of the circle?

Solution Equation of the circle?

### Example

x

Center, r = ?

Solution ^{2}+ y^{2}- 4x + 10y + 20 = 0Center, r = ?

x

x

(x

(x - 2)

(x - 2)

(x - 2)

Center: (2, -5), r = 3

^{2}+ y^{2}- 4x + 10y + 20 = 0x

^{2}- 4x + y^{2}+ 10y = -20(x

^{2}- 2⋅x⋅2 + 2^{2}) + (y^{2}+ 2⋅x⋅5 + 5^{2}) = -20 + 2^{2}+ 5^{2}- [1](x - 2)

^{2}+ (y + 5)^{2}= -20 + 4 + 25(x - 2)

^{2}+ (y + 5)^{2}= 9(x - 2)

^{2}+ (y - (-5))^{2}= 3^{2}Center: (2, -5), r = 3

[1]

Use (x

to make a perfect square trinomial.

x

→ x

y

→ y

+2

So, to undo the change,

write +2

^{2}- 4x) and (y^{2}+ 10y)to make a perfect square trinomial.

x

^{2}- 4x→ x

^{2}- 2⋅x⋅2 + 2^{2}y

^{2}+ 10x→ y

^{2}+ 2⋅x⋅5 + 5^{2}+2

^{2}and +5^{2}are added.So, to undo the change,

write +2

^{2}+ 5^{2}on the right side.Close

## Equation of a Tangent: Circle

### Formula

x

_{1}x + y

_{1}y = r

^{2}

### Example

Write an equation of the line

tangent to the circle x

Solution tangent to the circle x

^{2}+ y^{2}= 10 at (3, 1).## System of Equations: Circle, Line

### Example

x

y = x + 1

Solution ^{2}+ y^{2}= 25y = x + 1

x

y = x + 1

x

x

2x

x

(x + 4)(x - 3) = 0 - [3] [4]

1) x + 4 = 0

x = -4

y = -4 + 1 - [5]

= -3

(-4, -3)

2) x - 3 = 0

x = 3

y = 3 + 1 - [6]

= 4

(3, 4)

(-4, -3), (3, 4)

^{2}+ y^{2}= 25y = x + 1

x

^{2}+ (x + 1)^{2}= 25 - [1]x

^{2}+ x^{2}+ 2x + 1 = 25 - [2]2x

^{2}+ 2x - 24 = 0x

^{2}+ x - 12 = 0(x + 4)(x - 3) = 0 - [3] [4]

1) x + 4 = 0

x = -4

y = -4 + 1 - [5]

= -3

(-4, -3)

2) x - 3 = 0

x = 3

y = 3 + 1 - [6]

= 4

(3, 4)

(-4, -3), (3, 4)

Graph

[1]

[2]

[5]

x = -4 → y = x + 1

[6]

x = 3 → y = x + 1

Close