Aus Online Mathematik Brückenkurs 1

By completing the square, we can rewrite the x- and y-terms as quadratic expressions,

\displaystyle \begin{align} x^2 - 2x &= (x-1)^2 - 1^2\,,\\[5pt] y^2 + 6y &= (y+3)^2 - 3^2\,, \end{align}

and the whole equation then has standard form,

\displaystyle \begin{align} (x-1)^2 - 1 + (y+3)^2 - 9 &= -3\,,\\[5pt] (x-1)^2 + (y+3)^2 &= 7\,\textrm{.} \end{align}

From this, we see that the circle has its centre at (1,-3) and radius \displaystyle \sqrt{7}\,.