document.write( "Question 55199: Find the center and the radius of the circle whose equation is given: x^2+y^2+6x+8=0
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Algebra.Com's Answer #37461 by Earlsdon(6294) $\"\"$ $\"About$

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The general form of the equation for a circle with its center at (h, k) and radius of r is:
\n" ); document.write( " $\"%28x-h%29%5E2%2B%28y-k%29%5E2+=+r%5E2\"$ \r
\n" ); document.write( "\n" ); document.write( "Let's get your equation into this form, then we can find the center (h, k) and the radius (r). Starting with the given equation:
\n" ); document.write( " $\"x%5E2%2By%5E2%2B6x%2B8+=+0\"$ Complete the square in the x-terms by adding 1 to the 8 and 1 to the other side of the equation:
\n" ); document.write( " $\"%28x%5E2%2B6x%2B9%29+%2B+y%5E2+=+1\"$ Factor the parentheses.
\n" ); document.write( " $\"%28x%2B3%29%5E2+%2B+y%5E2+=+1\"$ Compare with the general form:
\n" ); document.write( " $\"%28x-h%29%5E2+%2B+%28y-k%29%5E2+=+r%5E2\"$ \r
\n" ); document.write( "\n" ); document.write( "You can see that the center (h, k) is (-3, 0) because h = -3 and k = 0 and the radius (r) is 1 because $\"r%5E2+=+1\"$ and $\"r+=+sqrt%281%29\"$ .
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