In order to find the center, you have to complete the square on both the x terms and the y terms. Start by moving the 48 over to the other side and grouping your like terms together, like this: [tex]( x^{2} -6x)+( y^{2} +16y)=-48[/tex]. Now complete the squares on the x and y terms. Take half of the linear term (6 for x and 16 for y), square it, and add it in to both sides. Like this: [tex]( x^{2} -6x+9)+( y^{2} +16y+64)=-48+9+64[/tex]. See what I did? I took half the linear term in the x group (half of 6 which is 3), then squared the 3 to get 9 and added it to both sides. Do that with the y grouping as well. This creates 2 perfect square binomials, one for the x terms and one for the y terms, which break down like this: [tex](x-3) ^{2} +(y+8) ^{2}=25 [/tex]. This puts the center of our circle at (3, -8) and the radius is 5
