Lösung 4.1:7a
Aus Online Mathematik Brückenkurs 1
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Version vom 14:48, 22. Okt. 2008
As the equation stands, it is difficult directly to know anything about the circle, but if we complete the square and combine x- and y-terms together in their own respective square terms, then we will have the equation in the standard form
| \displaystyle (x-a)^2 + (y-b)^2 = r^2\,, |
and we will then be able to read off the circle's centre and radius.
If we take the x- and y-terms on the left-hand side and complete the square, we get
| \displaystyle \begin{align}
x^2 + 2x &= (x+1)^2-1^2\,,\\[5pt] y^2 - 2y &= (y-1)^2-1^2\,, \end{align} |
and then the whole equation can be written as
| \displaystyle (x+1)^2 - 1^2 + (y-1)^2 - 1^2 = 1\,, |
or, with the constants moved to the right-hand side,
| \displaystyle (x+1)^2 + (y-1)^2 = 3\,\textrm{.} |
This is a circle having its centre at (-1,1) and radius \displaystyle \sqrt{3}\,.
