From Förberedande kurs i matematik 1

As the equation stands, it is difficult directly to know anything about the circle, but if we complete the square and combine \displaystyle x - and \displaystyle y - terms together in their own respective square terms, then we will have the equation in the standard form,

\displaystyle \left( x-a \right)^{2}+\left( y-b \right)^{2}=r^{2}

and we will then be able to read off the circle's centre and radius.

If we take the \displaystyle x - and \displaystyle y - terms on the left-hand side and complete the square, we get

\displaystyle x^{2}+2x=\left( x+1 \right)^{2}-1^{2}

\displaystyle y^{2}-2y=\left( y-1 \right)^{2}-1^{2}

and then the whole equation can be written as

\displaystyle \left( x+1 \right)^{2}-1^{2}+\left( y-1 \right)^{2}-1^{2}=1

or, with the constants moved to the right-hand side,

\displaystyle \left( x+1 \right)^{2}+\left( y-1 \right)^{2}=3

This is a circle having its centre at \displaystyle \left( -1 \right.,\left. 1 \right) and radius \displaystyle \sqrt{3}.