How can I determine the shape of a graph with $x^2$, $x$, and $y^2$ terms?

I'm working through a multivariable calculus course. I'm ~5 years removed from my most recent calculus course, and ~10 years removed from my most recent trigonometry course. While I'm understanding the new material pretty easily, I keep tripping up on material that I'm (understandably) expected to already know.

One review exercise is:

Draw the level curve for $f(x, y) = \frac{x}{x^2 + y^2}$ at values $c = -2, 0, 4$.

For $c = 0$, this is trivial ($x = 0, y \ne 0$). I'm having a bit more trouble with the other values. I can look in the back of the book and see that the shapes are two circles. But what signs am I looking for to tell me this?

I can simplify the equation of the level curve to something like this:

$$c = \frac{x}{x^2 + y^2}$$ $$c(x^2 + y^2) = x$$ $$cx^2 - x + cy^2 = 0$$ $$cy^2 = -cx^2 + x$$ $$y^2 = -x^2 + \frac{x}{c}$$ $$y = \pm \sqrt{-x^2 + \frac{x}{c}}$$

For $x \ne 0, c \ne 0$.

I also recall that equations like this describe a ellipse:

$$a(x - x_0)^2 + b(y - y_0)^2 = r^2$$

I assume there is a similar form that accounts for the possibility of an $x$ term. But I'm not sure how to go about determining this form.

This problem isn't assigned as homework, but I'm marking it as homework anyway, because it relates to a class that I'm taking.

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First, I think that you’re confusing circles with parabolas: replace parabola with circle throughout.

Here’s how I’d look at it. You have $cx^2-x+cy^2=0$. Divide through by $c$ and then complete the square in $x$ to get

$$\left(x-\frac1{2c}\right)^2+y^2=\frac1{4c^2}\;;$$

this is the equation of a circle with centre at $\left\langle\frac1{2c},0\right\rangle$ and radius $\frac1{2c}$. (That’s one of those things that you simply need to know; you probably should review the basic conics and their equations.)

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Thanks, this explanation makes perfect sense. More evidence of my rust: I had to google "complete the square" to refresh myself on the algorithm. I have a lot of review to do... – Matthew Pirocchi Sep 14 '12 at 21:18
@Matthew: Sorry: I actually meant to include a link with that. – Brian M. Scott Sep 14 '12 at 21:21
After working it through myself: Are you sure you don't mean $-\frac{1}{2c}$ and $\frac{1}{4c^2}$ in your final equation? – Matthew Pirocchi Sep 14 '12 at 21:31
@Matthew: I sure do; thanks for catching that. – Brian M. Scott Sep 14 '12 at 21:35