Question

x^{2} +y^{2} + 6y + 4x + 13 = 0

Answer 1

$x^2+y^2+6y+4x+13=0\ x^2+4x+4+y^2+6y+9=0\ (x+2)^2+(y+3)^2=0\ left { {{x+2=0} atop {y+3=0}} right. \ x=-2\ y=-3$