Solution 4.1:7b

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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\,.


Image:4_1_7_b.gif