Therefore the standard form of the circle is
[tex](x+1)^2+(y+\frac{1}{6})^2=1\frac{1}{36}[/tex]
Step-by-step explanation:
Given
[tex]3x^2+3y^2+6x-y=0[/tex]
[tex]\Leftrightarrow 3(x^2+y^2+2x-\frac{1}{3} y)=0[/tex]
[tex]\Leftrightarrow (x^2+2x+1) +(y^2+2.y.\frac{1}{6} +\frac{1}{36})-1- \frac{1}{36}=0[/tex]
[tex]\Leftrightarrow (x+1)^2+(y+\frac{1}{6})^2=1+\frac{1}{36}[/tex]
[tex]\Leftrightarrow (x+1)^2+(y+\frac{1}{6})^2=1\frac{1}{36}[/tex]
Therefore the standard form of the circle is
[tex](x+1)^2+(y+\frac{1}{6})^2=1\frac{1}{36}[/tex]
