Question

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

Answer 1

`(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