document.write( "Question 189073: So now that I have the standard form of the cricle x^2+y^2+2x-6y-6=0 wich is (x+3)^2+(y-3)^2=4^2 would the center be: -3, 3 and the radius: 4, or would it be 16? \n" ); document.write( "

Algebra.Com's Answer #141870 by Earlsdon(6294) $\"\"$ $\"About$

You can put this solution on YOUR website!
Hmmm...I don't quite see how you got your results from the given equation:
\n" ); document.write( "Given: $\"x%5E2%2By%5E2%2B2x-6y-6+=+0\"$ let's get this into the standard form for a circle: $\"%28x-h%29%5E2+%2B+%28y-k%29%5E2+=+r%5E2\"$ whose center is at (h, k) and whose radius is r.
\n" ); document.write( "Rearrange the given equation as shown below:
\n" ); document.write( " $\"%28x%5E2%2B2x%29%2B%28y%5E2-6y%29+=+6\"$ Now complete the square in x and in y.
\n" ); document.write( " $\"%28x%5E2%2B2x%2B1%29%2B%28y%5E2-6y%2B9%29+=+6%2B1%2B9\"$ Simplify.
\n" ); document.write( " $\"highlight%28%28x%2B1%29%5E2%2B%28y-3%29%5E2+=+16%29\"$ Compare this with the standard form above:
\n" ); document.write( " $\"%28x-h%29%5E2%2B%28y-k%29%5E2+=+r%5E2\"$ and you can see that: $\"h+=+-1\"$ and $\"k+=+3\"$ and $\"r+=+sqrt%2816%29\"$ , so...
\n" ); document.write( "The center is at: (-1, 3) and the radius is 4. \n" ); document.write( "

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