Aus Online Mathematik Brückenkurs 1

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

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

\displaystyle y^{2}+6y=\left( y+3 \right)^{2}-3^{2}

and the whole equation then has standard form,

\displaystyle \begin{align} & \left( x-1 \right)^{2}-1+\left( y+3 \right)^{2}-9=-3 \\ & \Leftrightarrow \quad \left( x-1 \right)^{2}+\left( y+3 \right)^{2}=7 \\ \end{align}

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