Rewrite the equation in the standard form of a circle: (x-h)^2+(y-k)^2=r^2

Question

The equation to rewrite is: 3x^2+3y^2+16x+14y-72=0

Answer 1

3(x^2+(16/3)x) + 3(y^2+(14/3)y) = 72

3(x^2 + (16/3)x + 64/9) + 3(y^2 + (14/3)y + 49/9) = 72 + 64/3 + 49/3

3(x+8/3)^2 + 3(y+7/3)^2 = 329/3

[x-(-8/3)]^2 + [y-(-7/3)]^2 = 329/9

[x-(-8/3)]^2 + [y-(-7/3)]^2 = [sqrt(329) / 3]^2