Answer
The equation given represents a circle of center $\Big(-1,\dfrac{1}{6}\Big)$ and radius $\dfrac{\sqrt{37}}{6}$
Work Step by Step
$3x^{2}+3y^{2}+6x-y=0$
Take out common factor $3$ from the left side of the equation:
$3\Big(x^{2}+y^{2}+2x-\dfrac{1}{3}y\Big)=0$
Take $3$ to divide the right side:
$x^{2}+y^{2}+2x-\dfrac{1}{3}y=\dfrac{0}{3}$
$x^{2}+y^{2}+2x-\dfrac{1}{3}y=0$
Group the terms with $x$ together and the terms with $y$ together:
$(x^{2}+2x)+\Big(y^{2}-\dfrac{1}{3}y\Big)=0$
Complete the square for each group by adding $\Big(\dfrac{b}{2}\Big)^{2}$ inside both parentheses and on the right side of the equation. For the first parentheses, $b=2$ and for the second parentheses, $b=-\dfrac{1}{3}$:
$\Big[x^{2}+2x+\Big(\dfrac{2}{2}\Big)^{2}\Big]+\Big[y^{2}-\dfrac{1}{3}y+\Big(-\dfrac{1}{3\cdot2}\Big)^{2}\Big]=0+\Big(\dfrac{2}{2}\Big)^{2}+\Big(-\dfrac{1}{3\cdot2}\Big)^{2}$
$(x^{2}+2x+1)+\Big(y^{2}-\dfrac{1}{3}y+\dfrac{1}{36}\Big)=0+1+\dfrac{1}{36}$
$(x^{2}+2x+1)+\Big(y^{2}-\dfrac{1}{3}y+\dfrac{1}{36}\Big)=\dfrac{37}{36}$
Factor the expressions inside the parentheses, which are perfect square trinomials:
$(x+1)^{2}+\Big(y-\dfrac{1}{6}\Big)^{2}=\dfrac{37}{36}$
The expression obtained is the equation of a circle in $(x-h)^{2}+(y-k)^{2}=r^{2}$ form,
where $\Big(-1,\dfrac{1}{6}\Big)$ is the center of the circle and $\dfrac{\sqrt{37}}{6}$ is its radius.
