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 y²- and y-terms into a quadratic term by completing the square
\displaystyle y^2 + 4y = (y+2)^2 - 2^2\,\textrm{.} |
After rewriting, the equation is
\displaystyle x^2 + (y+2)^2 = 4 |
and we see that the equation describes a circle having its centre at (0,-2) and radius \displaystyle \sqrt{4}=2\,.