if i am given an equation for circle how do i find its center

the equation of the circle is x(to the second power) + y(to the second power) - 4x + 2y =11

$$x^2+y^2-4x+2y=11\\\\ \underbrace{( x^2 - 4x )}_{=(x-2)^2-4} + \underbrace{(y^2 +2y )}_{=(y+1)^2-1} = 11 \\\\ (x-2)^2 +(y+1)^2 = 11 +1+4\\\\ (x \textcolor[rgb]{1,0,0}{ -2})^2 +(y \textcolor[rgb]{0,0,1}{+1})^2 = 16\\\\ \boxed{ Center_x = -(\textcolor[rgb]{1,0,0}{ -2}) = 2 }\\ \boxed { Center_y = -(\textcolor[rgb]{0,0,1}{ +1}) = -1 }$$

$$\\4x^2+2y^2=11\\\\ $This is not a circle. It is an ellipse$$ $.    

The centre is (0,0) because it is just x² and y² 

If, for example, the centre was    (3,-5)    then it would be (x-3)²   and   (y+5)²

Here it is :)

https://www.desmos.com/calculator/v35v5nrxpg

Here is an artical explaining elipses

http://www.mathopenref.com/coordgeneralellipse.html