Lösung 4.1:7b
Aus Online Mathematik Brückenkurs 1
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Version vom 14:48, 22. Okt. 2008
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\,.
