completing square for a circle In the following question: 

I don't understand how we can get from the original equation to the final equation using completing the square.
Any thoughts as how to get to the final equation?

$x + y = c(x^2 + y^2 + 1)$
$x^2 + y^2 - \frac xc - \frac yc + 1 = 0$
$x^2 - \frac xc + (\frac 1{2c})^2- (\frac 1{2c})^2 + y^2 - \frac yc + (\frac y{2c})^2- (\frac 1{2c})^2+ 1 = 0$
$(x - \frac 1{2c})^2 + (y-\frac 1{2c})^2 + 1 - \frac 1{4c^2} - \frac 1{4c^2} = 0$
$(x - \frac 1{2c})^2 + (y-\frac 1{2c})^2 = \frac 1{2c^2} -1$.
It's.... completing the square and exactly what it says.
