SOLUTION: Place in standard form for the circle ( which is (x-h)^2 + (y-k)^2 = r^2): x^2 + y^2 + 8x - 14y + 59 = 0

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: Place in standard form for the circle ( which is (x-h)^2 + (y-k)^2 = r^2): x^2 + y^2 + 8x - 14y + 59 = 0      Log On


   

Question 606646: Place in standard form for the circle ( which is (x-h)^2 + (y-k)^2 = r^2):
x^2 + y^2 + 8x - 14y + 59 = 0
Answer by ewatrrr(24785) About Me  (Show Source):
You can put this solution on YOUR website!
 

Hi,

x^2 + y^2 + 8x - 14y + 59 = 0

completing the respective Squares...

(x^2+8x +16) + (y^2 -14y + 49) -16-49 + 59 = 0

(x+4)^2 + (y-7)^2 = 6