The equation of a circle can be written in the standard form as follows: (x - h)\(^2\) + (y - k)\(^2\) = r\(^2\), where (h, k) is the center of the circle and r is its radius.

The given equation of the circle is x\(^2\) + y\(^2\) - 2\(\alpha\)x + 4y - \(\alpha\) = 0. We can rewrite this equation in the standard form by completing the square for both x and y terms:

x\(^2\) - 2\(\alpha\)x + y\(^2\) + 4y = \(\alpha\)

(x\(^2\) - 2\(\alpha\)x + \(\alpha\)\(^2\)) + (y\(^2\) + 4y + 4) = \(\alpha\) + \(\alpha\)\(^2\) + 4

(x - \(\alpha\))\(^2\) + (y + 2)\(^2\) = \(\alpha\)\(^2\) + \(\alpha\) + 4

So, the center of the circle is (\(\alpha\), -2) and its radius is \(\sqrt{\alpha^2+\alpha+4}\).

Since the radius is given to be 4 units, we have:

\(\sqrt{\alpha^2+\alpha+4}\) = 4

Squaring both sides, we get:

\(\alpha^2+\alpha+4\) = 16

Solving this quadratic equation, we get:

\(\alpha^2+\alpha-12=0\)

Factoring, we get:

(\(\alpha-3\))(\(\alpha+4\))=0

So, \(\alpha=3\) or \(\alpha=-4\).

Since we are looking for a positive value of \(\alpha\), we can conclude that \(\alpha=3\).