SOLUTION: Give the center of the circle with equation x squared+2x+y squared+22=0

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Question 90323This question is from textbook Algebra and Trigonometry
: Give the center of the circle with equation x squared+2x+y squared+22=0 This question is from textbook Algebra and Trigonometry

Answer by Earlsdon(6294) About Me  (Show Source):
You can put this solution on YOUR website!
You would like to get your equation into the standard form for a circle:
%28x-h%29%5E2%2B%28y-k%29%5E2+=+r%5E2 where the center is at (h, k) and the radius is r.
Starting with:
x%5E2%2B2x%2By%5E2%2B22+=+0 First, complete the square in the x-terms by adding the square of half the x-coefficient (%282%2F2%29%5E2+=+1) to both sides.
%28x%5E2%2B2x%2B1%29%2B%28y%5E2%29%2B22+=+1 Now subtract 22 from both sides.
%28x%5E2%2B2x%2B1%29%2B%28y%5E2%29+=+-21 Factor the trinomial in x.
%28x%2B1%29%5E2%2B%28y%2B0%29%5E2+=+-21 Now compare with the standard form for a circle:
%28x-h%29%5E2%2B%28y-k%29%5E2+=+r%5E2 ...and you can see that h = -1 and k = 0
So the center is at: (-1, 0)
Notice that the radius,(sqrt%28-21%29) is an imaginary number!