document.write( "Question 10828: What is the center and the radius of a circle with the equation:
\n" ); document.write( "x^2 + 6x + y^2 - 8y - 11 = 0 \n" ); document.write( "

Algebra.Com's Answer #5714 by Earlsdon(6294) $\"\"$ $\"About$

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The goal here is to get the equation of a circle in the standard form:\r
\n" ); document.write( "\n" ); document.write( " $\"%28x-h%29%5E2+%2B+%28y-k%29%5E2+=+r%5E2\"$ This is the equation of a circle with radius r and center at (h, k).\r
\n" ); document.write( "\n" ); document.write( "The technique is the \"complete the square\" in x and y.\r
\n" ); document.write( "\n" ); document.write( " $\"x%5E2+%2B+6x+%2B+y%5E2+-+8y+-+11+=+0\"$ Isolate the x and y terms by adding 11 to both sides.\r
\n" ); document.write( "\n" ); document.write( " $\"%28x%5E2+%2B+6x%29+%2B+%28y%5E2+-+8y%29+=+11\"$ Now complete the squares in x and y. This process entails adding a constant term to the x-group and to the y-group so that when you factor the x-group and the y-group, you will have a binomial squared in x and a binomial squared in y.
\n" ); document.write( "The constant for the x-group is found by squaring one-half of the coefficient of the x-term. That's (6/2)^2 = 9. Similarly for the y-group, you square one-half of the coefficient of the y-term. That's (-8/2)^2 = 16.\r
\n" ); document.write( "\n" ); document.write( " $\"%28x%5E2+%2B+6x+%2B+9%29+%2B+%28y%5E2+-+8y+%2B+16%29+=+11+%2B+9+%2B+16\"$ Don't forget to add the same numbers to both sides of the equation.\r
\n" ); document.write( "\n" ); document.write( " $\"%28x%5E2+%2B+6x+%2B+9%29+%2B+%28y%5E2+-+8y+%2B+16%29+=+36\"$ Now factor.\r
\n" ); document.write( "\n" ); document.write( " $\"%28x+%2B+3%29%5E2+%2B+%28y+-+4%29%5E2+=+6%5E2\"$ Now it looks like the standard form of the equation for a circle with radius of 6 and center at (-3, 4).\r
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