The center of the circle with equation $x^2+y^2=8x-6y-20$ is the point $(x,y)$. What is $x+y$?

\($x^2+y^2=8x-6y-20$\)

Write as

x^2  -  8x  +  y^2   +  6y   =  -20            

Complete the square on x and y

x^2  - 8x  + 16  +  y^2  + 6y  +  9   =  -20  + 16 + 9

(x - 4)^2   +  (y  + 3)^2  =  5

The center is   (4, -3)  =  (x, y)

So   

x + y  =   4  -  3     =   1