To find the center and radius of a circle and then graph it, the easiest way is to get the equation into standard form if it isn't already. The standard form for an equation of a circle is:

(x - h)² + (y - k)² = r²

       ... where (h, k) is the center and r is the radius.

So lets begin with an equation that is already in standard form. We will find the center and radius and then graph it.

(x - 5)² + (y + 3)² = 4

From the description above, can you tell what the center and radius is? Check below.

The center should be (5, -3) because you take the opposite of the number in each of the parantheses. This represents (h,k). To find r we take 4 and set it equal to r² since that is what it represents in the equation, take the square root to get 2 for the radius. We then use this information to graph the circle. See below.

First we plot the center. Then we count 2 up, down, right and left from the center and this gives a fairly accurate graph of the circle. If we want more accuracy, we must create an x,y chart to get some more points. Here is the graph of the circle.

If the equation of the circle is not in standard form, we must first get it into standard form by grouping the variables and completing the square where necessary. If there is a variable squared without a middle term of the same variable, it is unnecessary to complete the square on that variable only and then the coordinate for that variable will be 0. For example take the following equation to see what we mean.

x² + y² - 6y+ 2 = 0

Since the y's are already grouped together, the first step is move the 2 to the other side of the equation.

x² + y² - 6y + ____= - 2 + ____

x² + y² - 6y + 9 = - 2 + 9

     (take the middle coefficient, divide it by 2, and square it)

x² + (y - 3)² = 7

      ( factor the left and simplify the right)

Now we should be able to find the center and radius and then graph it like we did the first circle above. Check below.

The center is (0, 3) and the radius is √7 which for graphing purposes we approximate it to be a little over 2 but smaller than 3 since it is between the √4 and the √9. The graph of the circle is:

Lets take an equation where we have to complete the square on both variables. Look at the equation: 4x² + 4y² - 16x - 24y + 36 = 0.

Again we have to complete the square, so try it and check down below.

Group variables: 

     4x² -16x +___ + 4y² - 24y + ____ = - 36 + ____+____

Factor out the coefficient:

      4(x² - 4x + ___) + 4( y² - 6y + ____) = -36 + ____+_____

Divide middle term by 2 and square:

      4(x² - 4x + 4) + 4(y² - 6y + 9) = -36 +16 + 36

Factor the right side and simplify the left side:

      4(x - 2)² + 4(y - 3)² = 16

Divide by 4 to get it into the correct format of a circle: 

      (x-2)² + (y-3)² = 4

Now you should be able to give the center and radius and graph it. Check below.

The center is (2, 3) and the radius is 2. Now use this to graph the circle below: