x²+y²+14x+2y+14=0
In order to solve this kind of problem, you will have to go back to te standard equation of the circle: (x-h)² + (y-k)² = R², where h and k are the coordinate of the center and R the radius
To find this equation you have to complete the square as shown below:
a) Assemble all x together as well as the y's : (x² + 14x) + (y² + 2y) + 14 = 0
b) complete the square of (x²+14x) it gives: (x+7)² - 49
c) complete the square of (y² 2y) it gives: (y+1)² - 1
Rewrite the original equation:
(x+7)² - 49 + (y+1)² - 1 +14 =0; Or (x+7)² + (y+1)² = -14 +1 + 49
Finally the original equation rewritten in the standard form is:
: (x+7)² + (y+1)² = 36. So the coordinates of the center O (-7, -1)
