SOLUTION: 3x^2+3y^2−12x−12y−3=0 is the equation of a circle with center (h,k) and radius r for: h = k = r =

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: 3x^2+3y^2−12x−12y−3=0 is the equation of a circle with center (h,k) and radius r for: h = k = r =      Log On


   

Question 661577: 3x^2+3y^2−12x−12y−3=0 is the equation of a circle with center (h,k) and radius r for:
h =
k =
r =
Answer by lwsshak3(11628) About Me  (Show Source):
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3x^2+3y^2−12x−12y−3=0 is the equation of a circle with center (h,k) and radius r :
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standard form of equation for a circle:(x-h)^2+(y-k)^2=r^2, (h,k)=(x,y) coordinates of center, r=radius
3x^2+3y^2−12x−12y−3=0
complete the square
3x^2−12x+3y^2−12y−3=0
3(x^2−4x+4)+3(y^2−4y+4)=3+12+12
3(x-2)^2+3(y-2)^2=27
divide by 3
(x-2)^2+(y-2)^2=9
h =2
k =2
r =3