The center and radius for the circle $x^2 + y^2 +6x-4y +4 = 0$ respectively are: 

1. (2, 3) and 3 

2. (3, 2) and 8 

3. (2, -3) and 3 

4. (-3, 2) and 3

Solution

Equation: $$x^2 + y^2 + 6x - 4y + 4 = 0$$

Step 1: Group terms.
$$(x^2 + 6x) + (y^2 - 4y) + 4 = 0$$

Step 2: Complete the square.
- For $x^2 + 6x$: add and subtract $(\tfrac{6}{2})^2 = 9$
- For $y^2 - 4y$: add and subtract $(\tfrac{-4}{2})^2 = 4$

$$(x^2 + 6x + 9) + (y^2 - 4y + 4) + 4 - 9 - 4 = 0$$ $$\Rightarrow (x+3)^2 + (y-2)^2 - 9 = 0$$ $$\Rightarrow (x+3)^2 + (y-2)^2 = 9$$

Center: (-3, 2)

Radius: 3

Answer: Option 4