Given the equation of a circle as \(x^2 + y^2 - 6x + 8y + 9 = 0\), find the vertex form of the circle.

Step 1 ：The general form of a circle's equation is \(x^2 + y^2 + 2gx + 2fy + c = 0\), where the center of the circle is at \(-g, -f\) and the radius is \(r = \sqrt{g^2 + f^2 - c}\).

Step 2 ：Comparing this with the given equation, we see that \(g = -3\), \(f = 4\), and \(c = 9\).

Step 3 ：Substituting these values, we find that the center of the circle is at \(3, -4\) and the radius is \(r = \sqrt{(-3)^2 + (-4)^2 - 9} = \sqrt{9 + 16 - 9} = \sqrt{16} = 4\).

Step 4 ：So, the vertex form of the circle's equation is \((x - h)^2 + (y - k)^2 = r^2\), where \(h, k\) is the center of the circle and \(r\) is the radius.

Step 5 ：Substituting the values we found, we get \((x - 3)^2 + (y + 4)^2 = 4^2\).