# How to Graph a Circle Given a General or Standard Equation

*Ray is a Licensed Engineer in the Philippines. He loves to write any topic about mathematics and civil engineering.*

## What Is a Circle?

A circe is a locus of a point that moves such that it is always equidistant from a fixed point called the center. The constant distance is called the radius of the circle (r). The line joining the center of a circle to any points on the circle is known as the radius. The radius is an important measure of a circle because other measurements such as circumference and area can be determined if the measure of the radius is known. Being able to identify the radius can also be helpful in graphing the circle in the Cartesian Coordinate System.

### General Equation of a Circle

The general equation of a circle is where A = C and have the same sign. The general equation of a circle is either of the following forms.

**Ax**^{2}**+ Ay**^{2}**+ Dx + Ey + F = 0****x**^{2}**+ y**^{2}**+ Dx + Ey + F = 0**

To solve a circle, either one of the following two conditions must be known.

1. Use the general form of the circle when three points (3) along the circle are known.

2. Use the standard equation of the circle when the center (h,k) and the radius (r) are known.

### Standard Equation of a Circle

The left graph shows the equation and graph of the circle with center at (0,0) while the right graph shows the equation and graph of the circle with center at (h,k). For a circle with a form Ax^{2} + Ay^{2} + Dx + Ey + F = 0, the center (h,k) and radius (r) can be obtained using the following formulas.

**h = - D / 2A**

**k = - E / 2A**

**r = √[(D**^{2}** + E**^{2}** - 4AF)/4A**^{2}**]**

## Example 1

Graph and find the properties of a circle given the general equation x^{2} - 6x + y^{2} - 4y - 12 = 0.

### Solution

a. Convert the general form of the circle to standard form by completing the square.

x^{2} - 6x + y^{2} - 4y - 12 = 0

x^{2} - 6x + 9 + y^{2} - 4y + 4 = 12 + 9 + 4

(x - 3)^{2} + (y - 2)^{2} = 25

**Center (h,k) = (3,2)**

b. Solve for the radius of the circle from the standard equation of the circle.

(x - 3)^{2} + (y - 2)^{2} = 25

r^{2} = 25

**r = 5**

**Final Answer**: The center of the circle is at (3,2) and has a radius of 5 units.

## Example 2

Graph and find the properties of a circle given the general equation 2x^{2} + 2y^{2} - 3x + 4y - 1 = 0.

### Solution

a. Convert the general form of the circle to standard form by completing the square.

2x^{2} + 2y^{2} - 3x + 4y - 1 = 0

2(x^{2} - 3x/2 + 9/16) + 2(y^{2} + 2y + 1) = 1 + 2(9/16) + 2(1)

2(x - 3/2)^{2} + 2(y + 2)^{2} = 33/8

(x - 3/2)^{2} + (y + 2)^{2} = 33/16

**Center (h,k) = (3/2,-2)**

b. Solve for the radius of the circle from the standard equation of the circle.

(x - 3/2)^{2} + (y + 3)^{2} = 33/16

r^{2} = 33/16

**r = (√33)/4 units = 1.43 units**

**Final Answer**: The center of the circle is at (3/2,-2) and has a radius of 1.43 units.

## Example 3

Graph and find the properties of a circle given the general equation 9x^{2} + 9y^{2} = 16.

### Solution

a. Convert the general form of the circle to standard form by completing the square.

9x^{2} + 9y^{2} = 16

x^{2} + y^{2}= (4/3)^{2}

**Center (h,k) = (0,0)**

b. Solve for the radius of the circle from the standard equation of the circle.

x^{2} + y^{2}= (4/3)^{2}

**r = 4/3 units**

**Final Answer**: The center of the circle is at (0,0) and has a radius of 4/3 units.

## Example 4

Graph and find the properties of a circle given the general equation x^{2} + y^{2} - 6x + 4y - 23 = 0.

### Solution

a. Convert the general form of the circle to standard form by completing the square.

x^{2} + y^{2} - 6x + 4y - 23 = 0

(x^{2} - 6x + 9) + (y^{2} + 4y + 4) = 23 + 9 + 4

(x - 3)^{2} + (y + 2)^{2} = 36

**Center (h,k) = (3,-2)**

b. Solve for the radius of the circle from the standard equation of the circle.

(x - 3)^{2} + (y + 2)^{2} = 36

r^{2} = 36

**r = 6 units**

**Final Answer**: The center of the circle is at (3,-2) and has a radius of 6 units.

## Example 5

Graph and find the properties of a circle given the general equation x^{2} + y^{2} + 4x + 6y - 23 = 0.

**Solution**

a. Convert the general form of the circle to standard form by completing the square.

x^{2} + y^{2} + 4x + 6y - 23 = 0

x^{2} + 4x + 4 + y^{2} + 6y + 9 = 23 + 4 + 9

(x + 2)^{2} + (y + 3)^{2} = 36

**Center (h,k) = (-2,-3)**

b. Solve for the radius of the circle from the standard equation of the circle.

(x + 2)^{2} + (y + 3)^{2} = 36

r^{2} = 36

**r = 6 units**

**Final Answer**: The center of the circle is at (-2,-3) and has a radius of 6 units.

## Example 6

Find the radius and center of the circle given the general equation (x - 9/2)^{2} + (y + 2)^{2} = (17/2)^{2} and graph the function.

**Solution**

a. The given equation is already in standard form and there is no need to perform completing the square.

(x - 9/2)^{2} + (y + 2)^{2} = (17/2)^{2}

**Center (h,k) = (9/2,-2)**

b. Solve for the radius of the circle from the standard equation of the circle.

(x - 9/2)^{2} + (y + 2)^{2} = (17/2)^{2}

**r = 17/2 units = 8.5 units**

**Final Answer**: The center of the circle is at (9/2,-2) and has a radius of 8.5 units.

## Example 7

Find the radius and center of the circle given the general equation x^{2 }+ y^{2} + 6x - 14y + 49 = 0 and graph the function.

**Solution**

a. Convert the general form of the circle to standard form by completing the square.

x^{2 }+ y^{2} + 6x - 14y + 49 = 0

x^{2 }+ 6x + 9 + y^{2} - 14y + 49 = 32

(x + 3)^{2} + (y - 7)^{2} = 32

**Center (h,k) = (-3,7)**

b. Solve for the radius of the circle from the standard equation of the circle.

(x + 3)^{2} + (y - 7)^{2} = 32

**r = 5.66 units**

**Final Answer**: The center of the circle is at (-3,7) and has a radius of 5.66 units.

## Example 8

Find the radius and center of the circle given the general equation x^{2 }+ y^{2} + 2x - 2y - 23 = 0 and graph the function.

**Solution**

a. Convert the general form of the circle to standard form by completing the square.

x^{2 }+ y^{2} + 2x - 2y - 23 = 0

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

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

**Center (h,k) = (-1,1)**

b. Solve for the radius of the circle from the standard equation of the circle.

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

**r = 5 units**

**Final Answer**: The center of the circle is at (-1,1) and has a radius of 5 units.

## Learn How to Graph Other Conic Sections

- Graphing a Parabola in a Cartesian Coordinate System

The graph and location of a parabola depend on its equation. This is a step-by-step guide in graphing different forms of a parabola in the Cartesian coordinate system. - How to Graph an Ellipse Given an Equation

Learn how to graph an ellipse given the general form and standard form. Know the different elements, properties, and formulas necessary in solving problems about ellipse.

**© 2019 Ray**