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Complete the square to rewrite the equation into standard form, show all work please x^2+6x+y^2-4y=-9

User Wollnyst
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(x+3)^(2)+(y-2)^(2)=4

1) The first step is to divide the coefficient b by 2


\begin{gathered} b=6\therefore(6)/(2)=3 \\ b=-(4)/(2)\therefore=-(4)/(2)=-2 \end{gathered}

2) So now, let's add to both sides those squared numbers to both sides of the equation and then write them as binomials:


\begin{gathered} x^2+6x+3^2+y^2-4x+(-2)^2=-9+9+4 \\ (x+3)^2+(y-2)^2=4 \end{gathered}

Note that this is the Equation of the Circle in its standard form, whose radius is 4

User Donny Winston
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