Lösung 4.1:7a

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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

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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

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and then the whole equation can be written as

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or, with the constants moved to the right-hand side,

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This is a circle having its centre at (-1,1) and radius \displaystyle \sqrt{3}\,.


Image:4_1_7a-2(2).gif