document.write( "Question 8724: what is the center of a circle with the equation \"x%5E2%2By%5E2-4x%2B6y%2B1=0\" \n" ); document.write( "
Algebra.Com's Answer #4820 by FSAN85(6)\"\" \"About 
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OKay, this is what to do
\n" ); document.write( "first rearrange the equation:
\n" ); document.write( "x^2 - 4x + y^2 + 6y +1 = 0
\n" ); document.write( "take the 1 to ther side
\n" ); document.write( "x^2 - 4x + y^2 + 6y = -1
\n" ); document.write( "now expand ur equation
\n" ); document.write( "(x^2 - 4x +4) + (y^2 + 6y +9) = -1 + 4 + 9
\n" ); document.write( "break down
\n" ); document.write( "(x - 2)^2 + (y + 3)^2 = 12
\n" ); document.write( "Take the radical form of all the parts
\n" ); document.write( "(x-2) + (y+3) = (square root)12
\n" ); document.write( "set the x ans ys to 0, that will give u ur center which turn out to be:
\n" ); document.write( "A circle with a center at ( 2 , -3 ) with a radius 3.46 (squareroot of 12)
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