document.write( "Question 187674This question is from textbook Precalculus
\n" ); document.write( ": Hi,\r
\n" ); document.write( "\n" ); document.write( "3x^2 + 3y^2 - 6x - 6 = 0 (describe and sketch the given equation)\r
\n" ); document.write( "\n" ); document.write( "I believe this is a circle and I've tried \"completing the square\", but without the b(y) part of the equation, I'm lost. Please help.\r
\n" ); document.write( "\n" ); document.write( "Thank you,
\n" ); document.write( "Lynn
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Algebra.Com's Answer #140704 by stanbon(75887)\"\" \"About 
You can put this solution on YOUR website!
3x^2 + 3y^2 - 6x - 6 = 0 (describe and sketch the given equation)
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\n" ); document.write( "Divide thru by 3 to get:
\n" ); document.write( "x^2 + y^2 - 2x - 2 = 0
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\n" ); document.write( "x^2 - 2x + ? + y^2 = 2\r
\n" ); document.write( "\n" ); document.write( "x^2 - 2x + 1 + y^2 = 3\r
\n" ); document.write( "\n" ); document.write( "(x-1)^2 + (y-0)^2 = 3
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\n" ); document.write( "Circle with center at (1,0) and radius = sqrt(3)
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\n" ); document.write( "Cheers,
\n" ); document.write( "Stan H.
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