SOLUTION: x^2+y^2-16x+4Y+59=0 what is this put in standard form and graphed

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: x^2+y^2-16x+4Y+59=0 what is this put in standard form and graphed      Log On


   

Question 38917: x^2+y^2-16x+4Y+59=0 what is this put in standard form and graphed
Answer by fractalier(6550) About Me  (Show Source):
You can put this solution on YOUR website!
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.