From Förberedande kurs i matematik 1

We rewrite the equation in standard form by completing the square for the x- and y-terms,

\displaystyle \begin{align} x^{2} - 2x &= (x-1)^2 - 1^2\,,\\[5pt] y^{2} + 2y &= (y+1)^2 - 1^2\,\textrm{.} \end{align}

Now, the equation is

\displaystyle \begin{align} (x-1)^2 - 1 + (y+1)^2 - 1 &= -2\\ \Leftrightarrow\quad (x-1)^2 + (y+1)^2 &= 0\,\textrm{.} \end{align}

The only point which satisfies this equation is \displaystyle (x,y) = (1,-1) because, for all other values of x and y, the left-hand side is strictly positive and therefore not zero.