The center of the circle with equation \(x^2+y^2=4x+12y-39\) is the point (h,k) . What is h+k?
Ohhh I can do this one!
\(x^2+y^2=4x+12y-39 \\ x^2-4x+y^2-12y=-39 \\ x^2-4x+4+y^2-12y+36=-39+4+36 \\ (x-2)(x-2)+(y-6)(y-6)=1 \\ (x-2)^2 + (y-6)^2 = 1\)
So the center is located at (2,6)
2 + 6 = 8 :)
Bingo! Nice!
+4609 
