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