document.write( "Question 1066044: Find the standard form of the equation for the circle with the following properties. Endpoints of a diameter are (8,4) and (12,-4) \n" ); document.write( "
Algebra.Com's Answer #681210 by josgarithmetic(39617)\"\" \"About 
You can put this solution on YOUR website!
Center is the midpoint of the diameter.
\n" ); document.write( "\"x=%288%2B12%29%2F2=10\"
\n" ); document.write( "\"y=%284%2B%28-4%29%29%2F2=0\"
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\n" ); document.write( "\n" ); document.write( "Radius SQUARED is \"cross%28%288-12%29%5E2%2B%284-%28-4%29%29%5E2=16%2B64=80%29\" \"%281%2F2%29%5E2%2Asqrt%28%2812-8%29%5E2%2B%28-4-4%29%5E2%29%5E2\"\r
\n" ); document.write( "\n" ); document.write( "\"%281%2F4%29%28sqrt%2816%2B64%29%29%5E2\"\r
\n" ); document.write( "\n" ); document.write( "\"%281%2F4%2980\"\r
\n" ); document.write( "\n" ); document.write( "\"20\"\r
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\n" ); document.write( "\n" ); document.write( "EQUATION: \"%28x-10%29%5E2%2By%5E2=20\"
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