Answer:
The center of the circle is at the point (-3, -2) on the plane
Step-by-step explanation:
The first four steps in the completing of the squares to find the standard equation of a circle are correct. Let's start from there and actually complete the squares correctly:
[tex](x^2+6\,x+9)+(y^2+4\,y+4)=3+9+4\\(x+3)^2+(y+2)^2=16\\(x+3)^2+(y+2)^2=4^2[/tex]
Therefore, this is a circle centered at x = -3 [based on the horizontal translation defined by (x+3)], and at y = -2 [based on the vertical translation defined by (y+2)], and the circle has radius 4 based on the numerical constant squared on the right side of the equal sign.
