Equation of Circles
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Let's review what we already know about circles.

Definition:  A circle is a locus (set) of points in a plane equidistant from a fixed point.

Circle whose center is at the origin

Circle whose center is at (h,k)
(This will be referred to as the "center-radius form".
It may also be referred to as "
standard form".)

Example:  Circle with center (0,0), radius 4

Example:  Circle with center (2,-5), radius 3

Now, if we "multiply out" the above example we will get:

When multiplied out, we obtain the
"general form" of the equation of a circle.  Notice that in this form we can clearly see that the equation of a circle has both x2 and y2 terms and these terms have the same coefficient (usually 1).


When the equation of a circle appears in "general form", it is often beneficial to convert the equation to "center-radius" form to easily read the center coordinates and the radius for graphing.


1.  Convert   into center-radius form.

This conversion requires use of the technique of completing the square.

We will be creating two perfect square trinomials within the equation.

Start by grouping the x related terms together and the y related terms together.  Move any numerical constants (plain numbers) to the other side.
Get ready to insert the needed values for creating the perfect square trinomials.  Remember to balance both sides of the equation.
Find each missing value by taking half of the "middle term" and squaring.  This value will always be positive as a result of the squaring process.
Rewrite in factored form.

You can now read that the center of the circle is at (2, 3) and the radius is .



2How do the coordinates of the center of a circle relate to C and D when the equation of the circle is in the general form

Let's make some observations.  Re-examine our previous equations in general form and center-radius form.  Do you see a relationship between the center coordinates and C and D?

General form Center-radius form

C = -4,  D = 10

Center (2, -5)

C = -4,  D = -6

Center (2, 3)

It appears that the values of C and D are (-2) times the coordinates of the center respectively.  Why is this occurring?
When is expanded, becomes , where the center term's coefficient doubles the value of
-2.  Remember that while the equation deals with , the actual x-coordinate of the center of this circle is +2.

(Read more about these relationships at the Resource Page.)



3.  Write the equation of a circle whose diameter has endpoints (4, -1) and (-6, 7).
Find the center by using the midpoint formula.
Find the radius by using the distance formula.
Points (-6,7) and (-1,3) were used here.  (d = distance, or radius)



4.  Write the equation for the circle shown below if it is shifted 3 units to the right and 4 units up.

A shift of 3 units to the right and 4 units up places the center at the point (3, 4).  The radius of the circle can be seen from the graph to be 5.




5.  Convert into center-radius form.

Whoa!!!  This equation looks different.  Are we sure this is a circle???

In this equation, both the x and y terms appear in squared form and their coefficients (the numbers in front of them) are the same.  Yes, we have a circle here!  We will, however, have to deal with the coefficients of 2 before we can complete the square.

Center:  (-3/2, 2)        Radius:  1/2

  group the terms

  divide through by 2

  get ready to create perfect squares

  take half of the "middle term" and square it

  factor and write in center-radius form