Lösung 4.1:7b
Aus Online Mathematik Brückenkurs 1
The equation is almost in the standard form for a circle; all that is needed is for us to collect together the \displaystyle y^{\text{2}} - and \displaystyle y -terms into a quadratic term by completing the square
\displaystyle y^{2}+4y=\left( y+2 \right)^{2}-2^{2}
After rewriting, the equation is
\displaystyle x^{2}+\left( y+2 \right)^{2}=4
and we see that the equation describes a circle having its centre at
\displaystyle \left( 0 \right.,\left. -2 \right)
and radius
\displaystyle \sqrt{4}=2.
