SOLUTION: find the center of the circle defined by the equation 2x^2+2y^2+24x+28y+138=0. I'm kinda confused on how to start solving this problem, please help.

Algebra ->  Vectors -> SOLUTION: find the center of the circle defined by the equation 2x^2+2y^2+24x+28y+138=0. I'm kinda confused on how to start solving this problem, please help.      Log On


   

Question 372682: find the center of the circle defined by the equation 2x^2+2y^2+24x+28y+138=0.
I'm kinda confused on how to start solving this problem, please help.
Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
Complete the square in x and y.
2x%5E2%2B2y%5E2%2B24x%2B28y%2B138=0
x%5E2%2B12x%2By%5E2%2B14y%2B69=0
%28x%5E2%2B12x%2B36%29%2B%28y%5E2%2B14y%2B49%29%2B69=36%2B49
%28x%2B6%29%5E2%2B%28y%2B7%29%5E2=85-69
highlight%28%28x%2B6%29%5E2%2B%28y%2B7%29%5E2=16%29
Compare to the general equation for a circle centered at (h,k) with a radius R:
%28x-h%29%5E2%2B%28y-k%29%5E2=R%5E2