Question 38917
Let us rearrange and coomlete the square twice, so from
x^2 + y^2 - 16x + 4y + 59 = 0 we get
x^2 - 16x + 64 + y^2 + 4y + 4 = -59 + 64 + 4
(x - 8)^2 + (y + 2)^2 = 9
This is a circle with center at (8, -2) and radius 3.