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