SOLUTION: The center of a circle represented by the equation (x-2)^2+(y+3)^2=100 is located in what quadrant

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is a circle with center at (h,k) and radius r.

In your example, h = 2 and k = -3, so the center is at (2,-3).

The first quadrant is where x and y are both positive, and they are numbered counter-clockwise, so

Q1, x > 0, y > 0
Q2, x < 0, y > 0
Q3, x < 0, y < 0
Q4, x > 0, y < 0.

Your center is in Q4.