While the important technique of completing the square is on your fingertips,
it's a good time to discuss the equations of circles.
Suppose that a circle has center
$\,(h,k)\,$ and radius $\,r\,$.
Let $\,(x,y)\,$ be a typical point on this circle.
Refer to the diagram below.
The distance from $\,(x,y)\,$ to $\,(h,k)\,$ must equal $\,r\,$:
Thus, we have: 

For example, consider the equation
[beautiful math coming... please be patient]
$\,{(x3)}^2 + {(y+5)}^2 = 15\,$.
What does the graph of this equation look like?
That is, what does the set of all points $\,(x,y)\,$ that makes this equation true look like?
To answer, rename $\,5\,$ as $\,(5)\,$ and $\,15\,$ as [beautiful math coming... please be patient] $\,{(\sqrt{15})}^2\,$, and compare with the standard form of a circle: [beautiful math coming... please be patient] $$ \begin{alignat}{4} &{(x3)}^2 &&\quad+\quad &&{(y+5)}^2 &&\quad=\quad &15\cr &{(x3)}^2 &&\quad+\quad &&{(y(5))}^2 &&\quad=\quad &(\sqrt{15})^2\cr &{(xh)}^2 &&\quad+\quad &&{(yk)}^2 &&\quad=\quad &r^2 \end{alignat} $$
Thus,
[beautiful math coming... please be patient]
$\,h = 3\,$, $\,k = 5\,$, and $\,r = \sqrt{15}\,$.
The equation graphs as the circle with center $\,(h,k) = (3,5)\,$ and radius $\,r = \sqrt{15}\,$.
Students sometimes like to think of it this way:
You can use WolframAlpha to find the radius, center,
and graph of a circle. Just type in (say)
Unfortunately, circles aren't always given to you in this nice standard form.
Sometimes they're given to you all multiplied out.
In this situation, you should still be able to recognize that you're dealing with a circle,
and you should be able to put it in standard form.
For example, if you multiply out $\,(x3)^2 + (y+5)^2 = 15\,$, here's what you get:
[beautiful math coming... please be patient]
$$
x^2  6x + 9 + y^2 + 10y + 25 = 15
$$
Using the Addition and Multiplication Properties of Equality, it can look even different, say:
[beautiful math coming... please be patient]
$$
12x  2x^2 20y = 38 + 2y^2
$$
If the equation were originally given to you in this form, you should still be able
to recognize it as a circle!
(For fun, type this last equation into the WolframAlpha widget to convince yourself that
we're still dealing with the same circle.)
The process of identifying circles is actually quite easy, as discussed next.
Multiplying out the lefthand side of the equation $\,{(xh)}^2+{(yk)}^2=r^2\,$
gives:
[beautiful math coming... please be patient]
$$
x^2  2xh + h^2 + y^2  2ky + k^2 = r^2
$$
The key observation is that there are only five types of terms:
$\,x^2\,$, $\,y^2\,$, $\,x\,$, $\,y\,$, and constant.
And, when the $\,x^2\,$ and $\,y^2\,$ terms are on the same side of the equation,
then they will have the same coefficient.
These observations lead to the following way to recognize circles:
The following example illustrates the process of recognizing the equation of a circle
and putting it in standard form so that it can be easily graphed:
$3x^2  5x  7 = 1  3y^2$  original equation 
$3x^2  5x + 3y^2 = 8$  put all variable terms on the left, and constant terms on the right 
$\displaystyle x^2  \frac{5}{3}x + y^2 = \frac{8}{3}$  divide both sides by $\,3\,$; the coefficient of the squared term must be $\,1\,$ to use the technique of completing the square 
$\displaystyle x^2  \frac{5}{3}x + (\frac{5}{3\cdot2})^2 + y^2 = \frac{8}{3} + \frac{25}{36}$ 
There is an $\,x\,$ term, so we need to complete the square to get an expression of the form $\,(xh)^2\,$. On both sides, add $\,(\frac{5/3}{2})^2 = (\frac{5}{3\cdot 2})^2 = \frac{25}{36}\,$, which is the appropriate number to complete the square. 
$\displaystyle (x  \frac{5}{6})^2 + y^2 = \frac{121}{36} $  rename the perfect square trinomial; add fractions 
$\displaystyle (x\frac{5}{6})^2 + (y0)^2 = {(\frac{11}{6})}^2$  rename to make identification easier 
Once you know the center and radius of a circle, there are always four points that are easy to plot. Start at the center, and move up/down/left/right by the amount of the radius. Another way to gain confidence in your work is to check at least one of these four easy points in the original equation. Let's check the point $\,(\frac{5}{6},\frac{11}{6})\,$: $$ \begin{gather} 3x^2  5x  7 = 1  3y^2\cr 3\left(\frac{5}{6}\right)^2  5\left(\frac{5}{6}\right)  7 \ \ \overset{\text{? }}{=}\ \ 1  3\left(\frac{11}{6}\right)^2\cr \frac{109}{12} = \frac{109}{12}\qquad \text{Yes!} \end{gather} $$ 

To finish up, you may want to jump up to wolframalpha.com
and type in, say:
equation of circle with center (3,4) and radius square root of 2
Pretty amazing!
On this exercise, you will not key in your answer. However, you can check to see if your answer is correct. 
PROBLEM TYPES:
