Find the center and radius of the circular equation x^2 + y^2 - 4x + 6y + 9 = 0 ?

To find the center and radius of the circle given by the equation x^2 + y^2 - 4x + 6y + 9 = 0, we need to rewrite the equation in standard form.

First, complete the square for x and y terms:
x^2 - 4x + y^2 + 6y = -9
(x^2 - 4x + 4) + (y^2 + 6y + 9) = -9 + 4 + 9
(x - 2)^2 + (y + 3)^2 = 4

Now, the equation is in standard form (x - h)^2 + (y - k)^2 = r^2, where (h, k) is the center of the circle and r is the radius.

Comparing with the standard form, we can see that the center of the circle is at (2, -3) and the radius is √4 = 2.

Therefore, the center of the circle is (2, -3) and the radius is 2.

Step 1: Complete the square for both x and y terms.

```
x^2 - 4x + y^2 + 6y = -9
(x^2 - 4x + 4) + (y^2 + 6y + 9) = -9 + 4 + 9
(x - 2)^2 + (y + 3)^2 = 4
```

Step 2: Identify the center and radius.

The equation of a circle in standard form is:
```
(x - h)^2 + (y - k)^2 = r^2
```
where (h, k) is the center and r is the radius.

Comparing the given equation to the standard form, we have:
```
(x - 2)^2 + (y + 3)^2 = 4
```

Therefore, the center is (h, k) = (2, -3) and the radius is r = 2.