SOLUTION: 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?

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: 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?      Log On


   

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?
Found 2 solutions by kitty_tin, Earlsdon:
Answer by kitty_tin(5) About Me  (Show Source):
Answer by Earlsdon(6294) About Me  (Show Source):
You can put this solution on YOUR website!
Hmmm...I don't quite see how you got your results from the given equation:
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.
Rearrange the given equation as shown below:
%28x%5E2%2B2x%29%2B%28y%5E2-6y%29+=+6 Now complete the square in x and in y.
%28x%5E2%2B2x%2B1%29%2B%28y%5E2-6y%2B9%29+=+6%2B1%2B9 Simplify.
highlight%28%28x%2B1%29%5E2%2B%28y-3%29%5E2+=+16%29 Compare this with the standard form above:
%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...
The center is at: (-1, 3) and the radius is 4.