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 "centerradius
form".
It may also be referred to as "standard
form".) 
Equation:
Example:
Circle with center (0,0), radius 4
Graph:

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

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 x^{2}
and y^{2} 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 "centerradius" form to easily
read the center coordinates and the radius for graphing.
Examples:
1.
Convert
into centerradius 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
. 
2.
How 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.
Reexamine our previous equations in general form and centerradius
form. Do you see a relationship between the center coordinates and
C and D?
General form 
Centerradius 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 xcoordinate
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)

Equation: 
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.
Equation:

5.
Convert
into
centerradius 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 centerradius form

