document.write( "Question 24555: what is the center of the circle x^2 + 4x + y^2 + 2y - 9 = 0? \n" ); document.write( "

Algebra.Com's Answer #13072 by Earlsdon(6294) $\"\"$ $\"About$

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Find the centre of the circle: $\"x%5E2+%2B+4x+%2B+y%5E2+%2B+2y+-+9+=+0\"$
\n" ); document.write( "First get your equation into the standard form for a circle. You can do this by \"completing the square\" in the x-terms and in the y-terms.
\n" ); document.write( " $\"%28x%5E2+%2B+4x%29+%2B+%28y%5E2+%2B+2y+-+9%29+=+0\"$ Complete the square in the x- and y-terms by adding the square of half the x-, y-coefficients to both sides of the equation.
\n" ); document.write( " $\"%28x%5E2+%2B+4x+%2B+4%29+%2B+%28y%5E2+%2B+2y+%2B+1%29-+9+=+1%2B4\"$ Factor and Simplify.
\n" ); document.write( " $\"%28x+%2B+2%29%5E2+%2B+%28y+%2B+1%29%5E2+-+9+=+5\"$ Add 9 to both sides.
\n" ); document.write( " $\"%28x+%2B+2%29%5E2+%2B+%28y+%2B+1%29%5E2+=+14\"$ Now compare this with the standard form for a circle with centre at (h, k).
\n" ); document.write( " $\"%28x+-+h%29%5E2+%2B+%28y+-+k%29%5E2+=+r%5E2\"$
\n" ); document.write( "You can see that the centre of the circle is at (-2, -1) \n" ); document.write( "

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