This article is taken from the notes for my Math 107 web page. It’s purpose is to explain how to find equations for circles and also how to use this knowledge to provide a shortcut for graphing such equations. In order to do this you need to know two things.

- The distance formula
- The definition of a circle

For the distance formula see my notes.

#### What is a circle?

ã**A circle is a set of all points equidistant from a point.**

That is the standard geometry definition for a circle, and it is this kind of definition that allows us to write down equations for things like circles. Since we have a definition in terms of distances, we can use the distance formula to come up with an equation. Remember that the equation for a geometrical figure is an equation that all of the points in it have to satisfy, a kind of entrance test to get on the graph.

The point we want to make our points equidistant from is called the center and the distance is called the radius. (Think about how you draw a circle with a compass.) So let’s find the equation of the circle with radius r and center (h,k). I am using (h,k) for the coordinates of the circle rather than (x,y) because I have to reserve (x,y) for the points on the circle.

By the distance formula, the distance between (x,y) and (h,k) is given by this expression. By our definition of a circle, in order for (x,y) to pass the entrance test, this must be r, so we get this. This looks nicer if we square both sides, so we get this. for the standard equation of a circle with center (h,k) and radius r.

This means that if you were asked to write down the equation for a circle with a given center and radius you should be able to do this by just putting the numbers into this formula. You would substitute the x coordinate of the center for h and the y coordinate of the center for k and the radius for r.

**Example: **Find the equation of the circle with center (2,3) and radius 5.

**Solution: **Just put the 2 where the h is and the 3 where the k is and the 5 where the r is to get the equation.

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

This means that if someone were to plot all of the points that satisfy this equation, they would get exactly the circle that we want them to get, namely the one that has a center at (2,3) and a radius of 5.

Knowing about this standard equation for a circle can also be useful for making it easier to graph equations, because now whenever you see an equation like this, you don’t need to plot any points, you just need to recognize it is of this form and draw the circle. Before when you graphed equations you could only do it be just substituting x’s and y’s and finding points. If you have an equation of this kind of form and are asked to graph it, you need to find the center and the radius of the circle, so you need to recognize what the h, k, and r are. The right side of the equation will be r^{2}, so to find r, you have to take its square root. You need to put the left side into the form (x-h)^{2}+(y-k)^{2} in order to determine h and k. One problem you might have is that you might have addition instead of subtraction, so to deal with this you would have to see addition as subtracting a negative, which is kind of backwards of the way we normally think about these things. My suggestion for an approach so that you don’t have to remember too many things like that is just always find x and y so that what is in the parentheses is 0, and that will be your h and k, the coordinates of the center. Then once you have found the center and the radius of the circle, you can graph it by plotting the center and then drawing a circle of radius r around that center. A good way to do this is to plot four points, r to the right, r to the left, r up, and r down from the center and then just sketch a circle through those points. It doesn’t have to be perfect, just try your best. The main thing is that you show you teacher that you know what the center and radius is and what that means.

But since math instructors are all are mean and wicked, there is one more problem I need to tell how to deal with. The problem is, somebody just might have taken a nice equation for a circle like

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

and expanded everything out and rearranged things like this

x^{2}+2x+1+y^{2}-4y+4=9

x^{2}+y^{2}+2x-4y-5=0,

and then presented you with the equation

x^{2}+y^{2}+2x-4y-5=0

to graph. The problem is that it is a lot easier to expand that out than to put it back to where it came from. But there is a way to do it and it is to use the method of completing the squares. So here is what you do.

Get all the x’s together and all the y’s together and add something to both sides to get the constant on the right side of the equation. Leave space so that there is room to complete the square.

x^{2}+2x +y^{2}-4y =5

Then complete the square to find something to add to the x’s part and something to add to the y’s part so that they are both perfect squares. Remember to complete the square you take half of the coefficient and the square it, so for the x’s you get 2/2=1, 1^{2}=1, so you add 1, and for the y’s 4/2=2, 2^{2}=4, so you add 4. (You don’t have to worry about the minus on the 4 because you are going to square, so it will go away anyway.)

x^{2}+2x +1+y^{2}-4y +4=5+1+4

Then write the left side as squares and add up the right side and you get

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

and now you can find the center and radius and graph it. Here the center would be (-1,2) and the radius would be 3. So to graph it, all you need to do is find the point (-1,2) and then plot the points 3 up, 3 down, 3 to the right, and 3 to the left of it, and draw the circle through them.

by,

Erica Olivia Henry