document.write( "Question 8724: what is the center of a circle with the equation \n" );
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Algebra.Com's Answer #4820 by FSAN85(6)![]() ![]() ![]() You can put this solution on YOUR website! 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) \n" ); document.write( " |