Lesson Objectives
• Learn how to write the equation of a circle in center-radius form
• Learn how to sketch the graph of a circle
• Learn how to convert a circle in general form to center-radius form

## How to Graph a Circle

We have all seen a circle before, but let's give the circle an official definition. A circle is the set of all points in a plane that lie on a fixed distance from the center. The fixed distance is known as the radius of the circle.

### Equation of a Circle » Center-Radius Form

(x - h)2 + (y - k)2 = r2
This is the equation for a circle with a radius of r and a center of (h,k). Note that the radius is the principal square root of (r2). The negative square root is not considered since the radius is a distance and distance is non-negative.
To graph a circle whose equation is in center-radius form (standard form), we plot the center (h,k). We then plot four additional points using the radius. Starting at the center we move up, down, left, and right by the radius and plot each additional point. Lastly, we will draw a smooth curve through the outer four points. Let's look at an example.
Example 1: Sketch the graph of each circle
(x - 2)2 + (y + 4)2 = 9
First, let's write our equation in center-radius form. This will allow us to find the center and the radius of the circle.
(x - h)2 + (y - k)2 = r2
(x - 2)2 + (y - (-4))2 = 32
Since our center occurs at (h,k), we can state:
center: (2,-4)
This means we will plot the center of (2,-4) on the coordinate plane. Then we will use our radius to plot four additional points. Since the radius is 3, we will move from the center up by three units, down by three units, left by three units, and right by three units. This will give us the points:
(5,-4), (-1,-4), (2,-1), (2,-7)
We can plot these four points and connect them with a smooth curve.

### Completing the Square to Find the Center and Radius of a Circle

In some cases, we are not given the equation of a circle in standard form, otherwise known as center-radius form. When this happens, we can transform our equation by completing the square. Let's look at a few examples.
Example 2: Sketch the graph of each circle
x2 + y2 + 2x - 2y - 34 = 0
Let's group our x-terms and y-terms separately:
(x2 + 2x) + (y2 - 2y) - 34 = 0
Now complete the square for each. Recall completing the square means we are creating a perfect square trinomial. This means the trinomial can be factored into a binomial squared. To perform this action, we add 1/2 of the coefficient of the first-degree term (variable raised to the first power) squared to each side of the equation.
(x2 + 2x + 1) + (y2 - 2y + 1) - 34 = 2
We added 1 inside of the parentheses for each grouping. Since we added a total of 2 to the left side, we added 2 to the right side. Now we can factor each trinomial into a binomial squared.
(x + 1)2 + (y - 1)2 - 34 = 2
Now we can place this in center-radius form:
(x - (-1))2 + (y - 1)2 = 36
(x - (-1))2 + (y - 1)2 = 62
Since our center occurs at (h,k), we can state:
center: (-1, 1)
Additional points are found by moving 6 units up, down, left, and right from the center.
(-7, 1), (5, 1), (-1, 7), (-1, -5)
We can plot these four points and connect them with a smooth curve.

#### Skills Check:

Example #1

Find the center-radius form. $$x^2 + y^2 - 2x - 14y + 34=0$$

A
$$(x + 1)^2 + (y + 7)^2=16$$
B
$$(x - 1)^2 + (y - 7)^2=25$$
C
$$(x + 1)^2 + (y - 7)^2=16$$
D
$$(x - 1)^2 + (y - 7)^2=16$$
E
$$(x - 5)^2 + (y - 7)^2=25$$

Example #2

Find the center-radius form. $$x^2 + y^2 - 2x + 22y + 81=0$$

A
$$(x + 1)^2 + (y + 11)^2=41$$
B
$$x^2 + (y - 13)^2=41$$
C
$$x^2 + (y + 7)^2=13$$
D
$$(x - 10)^2 + (y + 3)^2=41$$
E
$$(x - 1)^2 + (y + 11)^2=41$$

Example #3

Match the graph to its equation.

A
$$(x - 3)^2 + (y - 4)^2=16$$
B
$$(x + 4)^2 + (y - 9)^2=3$$
C
$$(x + 4)^2 + (y + 3)^2=9$$
D
$$(x + 4)^2 + (y - 3)^2=9$$
E
$$(x - 4)^2 + (y - 3)^2=9$$