Let a line perpendicular to the line $2 x-y=10$ touch the parabola $y^2=4(x-9)$ at the point P. The distance of the point P from the centre of the circle $x^2+y^2-14 x-8 y+56=0$ is __________.

<p>Line perpendicular to $2 x-y=10$ have slope $=\frac{-1}{2}$</p> <p>$\Rightarrow$ Line tangent to parabola $y^2=4(x-9)$ with slope $m$ is</p> <p>$$\begin{aligned} & y=m(x-9)+\frac{1}{m}, m=\frac{-1}{2} \\ & \Rightarrow y=\frac{-(x-9)}{2}-2 \Rightarrow 2 y=-x+9-4 \\ & \Rightarrow 2 y+x=5 \end{aligned}$$</p> <p>Solving the tangent and parabola we get point $P$</p> <p>$$\begin{aligned} & \left(\frac{5-x}{2}\right)^2=4(x-9) \Rightarrow x^2-10 x+25=16 x-144 \\ & \Rightarrow x^2-26 x+169=0 \Rightarrow(x-13)^2=0 \\ & \Rightarrow P \equiv(13,-4) \end{aligned}$$</p> <p>Distance of $P$ from the centre of circle $(7,4)$ is $\sqrt{(13-7)^2+(-4-4)^2}=\sqrt{36+64}=10$ units.</p>