We have that the formula for the circle with a center of (-2, 3) and a radius of 3 is given by the formula:
[tex](x-(-2))^2+(y-3)^2=3^2=(x+2)^2+(y-3)^2=9[/tex]
Then, we can expand the previous expression as follows:
[tex]x^2+4x+4+y^2-6y+9=9[/tex]
Thus
[tex]x^2+4x+4+y^2-6y+9-9=0\Rightarrow x^2+4x+4+y^2-6y=0[/tex]
Therefore
[tex]x^2+y^2+4x-6y+4=0[/tex]
Then, the correct answer is option D.
