document.write( "Question 479503: Given the equation (x - 3)2 + y2 = 121 of a circle, identify the center and the radius. \n" ); document.write( "
Algebra.Com's Answer #328527 by Edwin McCravy(20056)\"\" \"About 
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document.write( "(x - 3)² + y² = 121\r\n" );
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document.write( "Ww want to first write it in the form \r\n" );
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document.write( "(x - h)² + (y - k)² = r², which has center (h,k) and radius r\r\n" );
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document.write( "(x - 3)² + y² = 121\r\n" );
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document.write( "The first term is already in the form (x - h)².  The second\r\n" );
document.write( "term is y², so to get it in the form (y - k)², we write it \r\n" );
document.write( "as (y - 0)²\r\n" );
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document.write( "(x - 3)² + (y - 0)² = 121\r\n" );
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document.write( "To get the last term in the form r² we write 121 as 11²:\r\n" );
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document.write( "(x - 3)² + (y - 0)² = 11²\r\n" );
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document.write( "Now we compare it to \r\n" );
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document.write( "(x - h)² + (y - k)² = r²\r\n" );
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document.write( "and get h = 3, k = 0, r = 11\r\n" );
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document.write( "So the center is (h,k) = (3,0) and the radius is r = 11\r\n" );
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document.write( "Edwin
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