Circles are useful in many Precalculus and Calculus problems (and in life).
For a thorough review, study the following two lessons.
Be sure to clickclickclick the web exercises in each lesson to check your understanding!
EQUATIONS OF CIRCLES
(standard form)
The equation of the circle with center $\,(h,k)\,$ and radius $\,r\,$ is:
[beautiful math coming... please be patient]
$$
(xh)^2 + (yk)^2 = r^2
$$
This form of the equation is often called the standard form of a circle.

For example, the circle with center $\,(2,3)\,$ and radius $\,5\,$ might make its appearance in any
of these
equivalent forms:
$(x  2)^2 + (y  (3))^2 = 5^2$  $(x2)^2 + (y+3)^2 = 25$  $x^2  4x + 4 + y^2 + 6y + 9 = 25$ 
$x^2  4x + y^2 + 6y = 12$  $x^2 + 6y = 12 + 4x  y^2$  $2x^2 + 12y = 24 + 8x  2y^2$ 
You need to be able to recognize any of these equations as a circle,
and then put it in standard form and graph it.
Here are the tools you need:
The technique of ‘completing the square’ is needed to put a circle in standard form:
$3x^2  5x  7 = 1  3y^2$  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 $\,(\frac{5}{6},\frac{11}{6})\,$: $$ \begin{gather} 3x^2  5x  7 = 1  3y^2\cr 3\left(\frac{5}{6}\right)^2  5\left(\frac{5}{6}\right)  7 \ \ \overset{\text{?}}{=}\ \ 1  3\left(\frac{11}{6}\right)^2\cr \frac{109}{12} = \frac{109}{12}\qquad \text{Yes!} \end{gather} $$ 

For fun, jump up to
WolframAlpha
and type in:
properties of 3x^2  5x  7 = 1  3y^2
How easy is that!?
On this exercise, you will not key in your answer. However, you can check to see if your answer is correct. 
PROBLEM TYPES:
