1. Find the slope of the given line

The line is $2x-y=10 \implies y=2x-10$ So its slope is $2$.

A line perpendicular to this has slope $m=-\frac{1}{2}$

2. Use the tangent form for the parabola

The parabola is $y^2=4(x-9)$ This is of the form $y^2=4a(x-h)$ with $a=1$ and vertex $(9,0)$.

For the standard parabola $y^2=4ax$, the tangent with slope $m$ is $y=mx+\frac{a}{m}$ For the shifted parabola $y^2=4(x-9)$, replace $x$ by $(x-9)$: $y=m(x-9)+\frac{1}{m}$

Now substitute $m=-\frac12$: $y=-\frac12(x-9)+\frac{1}{-1/2}$ $y=-\frac{x}{2}+\frac{9}{2}-2$ $y=-\frac{x}{2}+\frac{5}{2}$

So the tangent line is $y=-\frac{x}{2}+\frac{5}{2}$

3. Find the point of contact $P$

Substitute the tangent line into the parabola: $y=-\frac{x}{2}+\frac{5}{2} \implies x=5-2y$

Now use $y^2=4(x-9)$ $y^2=4((5-2y)-9)=4(-4-2y)=-16-8y$

Thus, $y^2+8y+16=0$ $(y+4)^2=0$ So, $y=-4$

Then $x=5-2(-4)=13$

Hence the point of contact is $P=(13,-4)$

4. Find the centre of the circle

The circle is $x^2+y^2-14x-8y+56=0$

Complete the squares: $x^2-14x+y^2-8y+56=0$ $(x-7)^2-49+(y-4)^2-16+56=0$ $(x-7)^2+(y-4)^2=9$

So the centre is $C=(7,4)$

5. Find the distance $CP$

$CP=\sqrt{(13-7)^2+(-4-4)^2}$ $=\sqrt{6^2+(-8)^2}$ $=\sqrt{36+64}$ $=\sqrt{100}=10$

6. Final answer

The required distance is $\boxed{10}$