completing square for a circle

Question

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?

Answer 1

$x + y = c(x^2 + y^2 + 1)$

$x^2 + y^2 - \frac xc - \frac yc + 1 = 0$

$x^2 - \frac xc + (\frac 1{2c})^2- (\frac 1{2c})^2 + y^2 - \frac yc + (\frac y{2c})^2- (\frac 1{2c})^2+ 1 = 0$

$(x - \frac 1{2c})^2 + (y-\frac 1{2c})^2 + 1 - \frac 1{4c^2} - \frac 1{4c^2} = 0$

$(x - \frac 1{2c})^2 + (y-\frac 1{2c})^2 = \frac 1{2c^2} -1$.

It's.... completing the square and exactly what it says.

Answer 2

One way of doing this is to avoid fractions completely until the last step.

$$\begin{align} \frac{x+y}{x^2+y^2+1}&=c\\ x+y&=c(x^2+y^2+1)\\ x+y&=cx^2+cy^2+c\\ cx^2-x+cy^2-y+c&=0\\ cx^2-x+cy^2-y&=-c\\ 4c(cx^2-x+cy^2-y)&=4c(-c)\\ 4c^2x^2-4cx+4c^2y^2-4cy&=-4c^2\\ (2cx)^2-2(2cx)+(2cy)^2-2(2cy)&=-4c^2\\ (2cx)^2-2(2cx)+1+(2cy)^2-2(2cy)+1&=2-4c^2\\ (2cx-1)^2 + (2cy-1)^2 &= 2-4c^2\\ \frac{(2cx-1)^2 + (2cy-1)^2}{4c^2} &= \frac{2-4c^2}{4c^2}\\ \bigg(x-\frac{1}{2c}\bigg)^2 + \bigg(y-\frac{1}{2c}\bigg)^2 &= \frac{1}{2c^2}-1\\ \end{align}$$