document.write( "Question 187674This question is from textbook Precalculus
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document.write( ": Hi,\r
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document.write( "3x^2 + 3y^2 - 6x - 6 = 0 (describe and sketch the given equation)\r
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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
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document.write( "Thank you,
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document.write( "Lynn \n" );
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Algebra.Com's Answer #140704 by stanbon(75887) ![]() You can put this solution on YOUR website! 3x^2 + 3y^2 - 6x - 6 = 0 (describe and sketch the given equation) \n" ); document.write( "------------------ \n" ); document.write( "Divide thru by 3 to get: \n" ); document.write( "x^2 + y^2 - 2x - 2 = 0 \n" ); document.write( "--- \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 \n" ); document.write( "-------------------------------- \n" ); document.write( "Circle with center at (1,0) and radius = sqrt(3) \n" ); document.write( "====================================================== \n" ); document.write( "Cheers, \n" ); document.write( "Stan H. \n" ); document.write( " |