document.write( "Question 298099: What is the equation in standard form for the circle: x^2+y^2-4x+8y+4=0 \n" ); document.write( "
Algebra.Com's Answer #214534 by Earlsdon(6294)\"\" \"About 
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Rewrite in standard form:
\n" ); document.write( "\"x%5E2%2By%5E2-4x%2B8y%2B4+=+0\" Group the x-terms and the y-terms as shown:
\n" ); document.write( "\"%28x%5E2-4x%29%2B%28y%5E2%2B8y%29%2B4+=+0\" Subtract 4 from both sides.
\n" ); document.write( "\"x%5E2-4x%29%2B%28y%5E2%2B8y%29+=+-4\" Now complete the square in x and y by adding the square of half the x-term coefficient and the square of half the y-term coefficient to both sides of the equation.
\n" ); document.write( " Simplify both sides of the equation.
\n" ); document.write( "\"%28x%5E2-4x%2B4%29%2B%28y%5E2%2B8y%2B16%29+=+-4%2B4%2B16\" Now factor the left side and simplify the right side.
\n" ); document.write( "\"%28x-2%29%5E2%2B%28y%2B4%29%5E2+=+16\" Finally...
\n" ); document.write( "\"highlight%28%28x-2%29%5E2%2B%28y%2B4%29%5E2+=+4%5E2%29\" Compare with standard form...
\n" ); document.write( "\"%28x%2Bh%29%5E2%2B%28y%2Bk%29%5E2+=+r%5E2\" and you can see that \"h+=+-2\",\"k+=+4\", and \"r+=+4\" and this represents a circle with center at (-2, 4) and radius of 4.
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