Answer:
To convert the given equation of the circle to standard form we should do as follows.
[tex]x^{2} +y^{2}+4x-6y-36+36=36\\ x^{2} +y^{2}+4x-6y=36[/tex]
[tex]x^{2} +4x +y^{2}-6y=36[/tex]
[tex]x^{2} +4x + (\frac{4}{2} )^{2} +y^{2} -6y+(\frac{6}{2} )^{2} =36[/tex]
[tex]x^{2} +4x+4+y^{2} -6y+9=36+4+9\\[/tex]
[tex](x+2)^{2} +(y-3)^{2} =49[/tex]
As you can observe, the circle has center at (-2,3) and it has a radius of 7 units.
Notice that the second choice is correct, because we added 4 units to both sides to complete the square for the x terms.