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

$\\$ `x^2+y^2+4y+13=6x`

$\\$ `<=>x^2-6x+9+y^2+4y+4=0`

$\\$ `<=>(x-3)^2+(y+2)^2=0`

$\\$ Vì `(x-3)^2>=0∀x`

$\\$     `(y+2)^2>=0∀y`

$\\$ `=>(x-3)^2+(y+2)^2>=0∀x;y`

$\\$ Dấu `=` xảy ra `<=>{(x-3=0),(y+2=0):}`

$\\$ `<=>{(x=3),(y=-2):}`

$\\$ Vậy `(x;y)=(3;-2)`