document.write( "Question 516748: Determine whether the equation represents a circle, a point, or has no graph. If the equation is that of a circle, find its center and radius. x^2+y^2+72=12x \n" ); document.write( "

Algebra.Com's Answer #344507 by Edwin McCravy(20056) $\"\"$ $\"About$

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document.write( "                x² + y² + 72 = 12x\r\n" );
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document.write( "If it has a graph at all it would be a circle since it has both\r\n" );
document.write( "x² and y² terms with the same coefficient when on the same side\r\n" );
document.write( "of the equation.  So we'll try to get it into the form of a\r\n" );
document.write( "circle, which is\r\n" );
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document.write( "             (x-h)² + (y-k)² = r²\r\n" );
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document.write( "                x² + y² + 72 = 12x\r\n" );
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document.write( "Rearrange by subtracting 12x from both sides,\r\n" );
document.write( "and put the subtacted 12x next to the x² term:\r\n" );
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document.write( "          x² - 12x + y² + 72 = 0\r\n" );
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document.write( "Subtract 72 from both sides:\r\n" );
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document.write( "               x² - 12x + y² = -72 \r\n" );
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document.write( "Complete the square by taking  of -12, the coefficient of x,\r\n" );
document.write( "which gives -6, then square -6 to get +36, and add 36 to both\r\n" );
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document.write( "          x² - 12x + 36 + y² = -72 + 36\r\n" );
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document.write( "Factor the first three terms, x² - 12x + 36, as (x - 6)(x - 6) which \r\n" );
document.write( "can be written (x - 6)²\r\n" );
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document.write( "               (x - 6)² + y² = -72 + 36\r\n" );
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document.write( "Write y² as (y - 0)² and combine the terms on the right\r\n" );
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document.write( "         (x - 6)² + (y - 0)² = -36\r\n" );
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document.write( "oh, oh!  There can be no graph because the terms on the left are\r\n" );
document.write( "squares which are both non-negative, making their sum non-negative, \r\n" );
document.write( "yet the term on the right is negative,  So this cannot be the equation \r\n" );
document.write( "of any graph.\r\n" );
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document.write( "Edwin

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