$x^{2}+y^{2}-8 x+16 y-1=0$ is the equation of a circle with center $(h, k)$ and radius $r$ for:
\[
h=
\]
and
\[
k=
\]
and
\[
r=
\]

Step 1 ：The given equation is \(x^{2}+y^{2}-8 x+16 y-1=0\).

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

Step 3 ：We can rewrite the given equation in this form to find the values of \(h\), \(k\), and \(r\).

Step 4 ：The correct form should be \((x-h)^2 + (y-k)^2 = r^2\), where \(h = -b/(2*a)\), \(k = -d/(2*a)\), and \(r = \sqrt{(b^2 + d^2 - 4*a*e)/(4*a^2)}\).

Step 5 ：After calculation, we find that the center of the circle is at \((h, k) = (-0.5, -8.0)\) and the radius of the circle is \(r = 8.08\).

Step 6 ：So, the final answer is: The center of the circle is at \(\boxed{(-0.5, -8.0)}\) and the radius of the circle is \(\boxed{8.08}\).