document.write( "Question 661577: 3x^2+3y^2−12x−12y−3=0 is the equation of a circle with center (h,k) and radius r for:\r
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Algebra.Com's Answer #411692 by lwsshak3(11628)\"\" \"About 
You can put this solution on YOUR website!
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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\n" ); document.write( "standard form of equation for a circle:(x-h)^2+(y-k)^2=r^2, (h,k)=(x,y) coordinates of center, r=radius
\n" ); document.write( "3x^2+3y^2−12x−12y−3=0
\n" ); document.write( "complete the square
\n" ); document.write( "3x^2−12x+3y^2−12y−3=0
\n" ); document.write( "3(x^2−4x+4)+3(y^2−4y+4)=3+12+12
\n" ); document.write( "3(x-2)^2+3(y-2)^2=27
\n" ); document.write( "divide by 3
\n" ); document.write( "(x-2)^2+(y-2)^2=9
\n" ); document.write( "h =2
\n" ); document.write( "k =2
\n" ); document.write( "r =3
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