SOLUTION: Give the center of the circle with equation x^2+2x+y^2+22=0 I’ve come up with the answer to be (-1,0) is this right?

Algebra ->  Circles -> SOLUTION: Give the center of the circle with equation x^2+2x+y^2+22=0 I’ve come up with the answer to be (-1,0) is this right?       Log On


   

Question 92916: Give the center of the circle with equation x^2+2x+y^2+22=0
I’ve come up with the answer to be (-1,0) is this right?
Found 2 solutions by psbhowmick, chitra:
Answer by psbhowmick(878) About Me  (Show Source):
You can put this solution on YOUR website!
x%5E2%2B2x%2By%5E2%2B22=0
%28x%2B1%29%5E2%2By%5E2%2B21=0
So centre is at (-1, 0).
You are right!

Answer by chitra(359) About Me  (Show Source):
You can put this solution on YOUR website!
The given equation of the circle is:

+x%5E2+%2B+2x+%2B+y%5E2+%2B+22+=+0


Rearranging terms and completing the square, we get:


x%5E2+%2B+2x+%2B+1+-+1+%2B+y%5E2+%2B+22+=+0+


+%28x+%2B+1%29%5E2+%2B+y%5E2+%2B+22+-+1+=+0+


+%28x+%2B+1%29%5E2+%2B+y%5E2+=+-21+


%28x+%2B+1%29%5E2+%2B+%28y+-+0%29%5E2+=+-21


Comparing the above equation with the standard eequation of the circle, we get:

+%28x+-+h%29%5E2+%2B+%28y+-+k%29%5E2+=+r%5E2


where (h,k) forms the centre and r the radius.


so here on comparing, we find (h,k) = (-1,0)


Thus the solution..

Regards