document.write( "Question 370517: Hi, I have to find the standard form of a circle with center (6, -3) and diameter \"20sqrt%282%29\".\r
\n" ); document.write( "\n" ); document.write( "Since the standard form of a circle is:\r
\n" ); document.write( "\n" ); document.write( "\"%28x+-+h%29%5E2+%2B+%28y+-+k%29%5E2+=+r%5E2\"\r
\n" ); document.write( "\n" ); document.write( "and diameter = \"r%5E2\",\r
\n" ); document.write( "\n" ); document.write( "\"%28x+-+6%29%5E2+%2B+%28y+%2B+3%29%5E2+=+20sqrt%282%29\"\r
\n" ); document.write( "\n" ); document.write( "I am getting this wrong and no clue why :( \r
\n" ); document.write( "\n" ); document.write( "Thanks for your help!
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Algebra.Com's Answer #264096 by jim_thompson5910(35256)\"\" \"About 
You can put this solution on YOUR website!
The left side is correct. The right side is where you're going wrong. Note: the diameter is \"2r\" and NOT \"r%5E2\"\r
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\n" ); document.write( "\n" ); document.write( "Since the diameter is \"d=20sqrt%282%29\", this means that the radius is half of this at \"r=d%2F2=20sqrt%282%29%2F2=10sqrt%282%29\"\r
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\n" ); document.write( "\n" ); document.write( "So \"r=10sqrt%282%29\"\r
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\n" ); document.write( "\n" ); document.write( "Now square 'r' to get \r
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\n" ); document.write( "\n" ); document.write( "So \"r%5E2=200\"\r
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\n" ); document.write( "\n" ); document.write( "This means that the equation of the circle is \"%28x-6%29%5E2%2B%28y%2B3%29%5E2=200\"\r
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\n" ); document.write( "\n" ); document.write( "If you need more help, email me at jim_thompson5910@hotmail.com\r
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\n" ); document.write( "\n" ); document.write( "Also, feel free to check out my tutoring website\r
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\n" ); document.write( "\n" ); document.write( "Jim
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