Answer
$${\text{circle}}$$
Work Step by Step
$$\eqalign{
& 3{x^2} + 6x + 3{y^2} - 12y = 12 \cr
& {\text{Factor and complete the square}} \cr
& \left( {3{x^2} + 6x} \right) + \left( {3{y^2} - 12y} \right) = 12 \cr
& 3\left( {{x^2} + 2x} \right) + 3\left( {{y^2} - 4y} \right) = 12 \cr
& \left( {{x^2} + 2x} \right) + \left( {{y^2} - 4y} \right) = 4 \cr
& \left( {{x^2} + 2x + 1} \right) + \left( {{y^2} - 4y + 4} \right) = 4 + 1 + 4 \cr
& {\left( {x + 1} \right)^2} + {\left( {y - 2} \right)^2} = 9 \cr
& {\text{The equation is in the form }}{\left( {x - h} \right)^2} + {\left( {y - h} \right)^2} = {r^2} \cr
& {\text{Therefore, the equation represents a circle centered at }}\left( { - 1,2} \right) \cr
& {\text{Graph}} \cr} $$
