Final answer:
The center of the circle with the equation x^2 - 2x + y^2 + 4y + 1 = 0 is (1, -2), and the radius is 2 after completing the square process and bringing the equation to its standard form. The correct answer is D) Center: (1, -2), Radius: 2.
Step-by-step explanation:
To find the center and the radius of the circle given by the equation x^2 - 2x + y^2 + 4y + 1 = 0, we first need to write the equation in the standard form of a circle. The standard form of a circle's equation is (x - h)^2 + (y - k)^2 = r^2, where (h, k) represents the circle's center and r is the radius.
Completing the square for the x-terms, we add and subtract (2/2)^2, which is 1, within the equation. Doing the same for the y-terms, we add and subtract (4/2)^2, which is 4, to complete the square. After adding 1 and 4 to both sides, the equation becomes:
(x^2 - 2x + 1) + (y^2 + 4y + 4) = 0 + 1 + 4
Now the equation can be rewritten as:
(x - 1)^2 + (y + 2)^2 = 4
Now we can easily see that the circle's center is (1, -2) and the radius is 2, because r^2 is equal to 4 and the radius r is the square root of r^2. Therefore, the correct answer is D) Center: (1, -2), Radius: 2.