The graph of the equation x^2+y^2+2y-2=0 can be described as a circle with a center at (0,-1).
To understand why, let's simplify the equation by completing the square.
First, rearrange the terms:
x^2 + (y^2 + 2y) = 2
Next, focus on the term involving y. To complete the square, we need to add and subtract the square of half of the coefficient of y, which is 1:
x^2 + (y^2 + 2y + 1 - 1) = 2
Simplifying this equation further, we have:
x^2 + (y+1)^2 - 1 = 2
Now, move the constant term to the right side of the equation:
x^2 + (y+1)^2 = 3
Comparing this equation to the standard form of a circle, (x-h)^2 + (y-k)^2 = r^2, we can determine that the center of the circle is at the point (h, k). In this case, the center is (0, -1).
Therefore, the correct answer is B. a circle with a center at (0, -1)