The center of the circle with equation \(x^2+y^2=8x-6y-20\) is the point (x,y) . What is x+y ?
\(x^2+y^2=8x-6y-20 \\ x^2-8x+y^2+6y=-20 \\ x^2-8x+16+y^2+6y+9=-20+16+9 \\ (x-4)(x-4)+(y+3)(y+3)=5 \\ (x-4)^2 + (y+3)^2 = 5\)
So the center is located at (4,-3)
4 + (-3) = 4 - 3 = 1
Bingo!
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