$x^2+y^2-8x+16y-1=0$ 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 $x^2+y^2-8x+16y-1=0$.

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 $h=-b/(2\ast a)$, $k=-d/(2\ast a)$, and $r=\sqrt{(b^2+d^2-4\ast a\ast e)/(4\ast a^2)}$.

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 $\boxed{{\displaystyle (-0.5,-8.0)}}$ and the radius of the circle is $\boxed{{\displaystyle 8.08}}$.