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?

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) (Show Source): You can put this solution on YOUR website!
r=4
Answer by Earlsdon(6294) (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: let's get this into the standard form for a circle: whose center is at (h, k) and whose radius is r.
Rearrange the given equation as shown below:
Now complete the square in x and in y.
Simplify.
Compare this with the standard form above:
and you can see that: and and , so...
The center is at: (-1, 3) and the radius is 4.
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