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 \displaystyle y^{\text{2}} - and \displaystyle y -terms into a quadratic term by completing the square


\displaystyle y^{2}+4y=\left( y+2 \right)^{2}-2^{2}


After rewriting, the equation is


\displaystyle x^{2}+\left( y+2 \right)^{2}=4


and we see that the equation describes a circle having its centre at \displaystyle \left( 0 \right.,\left. -2 \right) and radius \displaystyle \sqrt{4}=2.