Line perpendicular to $$2 x-y=10$$ have slope $$=\frac{-1}{2}$$

$$\Rightarrow$$ Line tangent to parabola $$y^2=4(x-9)$$ with slope $$m$$ is

$$\begin{aligned} & y=m(x-9)+\frac{1}{m}, m=\frac{-1}{2} \\ & \Rightarrow y=\frac{-(x-9)}{2}-2 \Rightarrow 2 y=-x+9-4 \\ & \Rightarrow 2 y+x=5 \end{aligned}$$

Solving the tangent and parabola we get point $$P$$

$$\begin{aligned} & \left(\frac{5-x}{2}\right)^2=4(x-9) \Rightarrow x^2-10 x+25=16 x-144 \\ & \Rightarrow x^2-26 x+169=0 \Rightarrow(x-13)^2=0 \\ & \Rightarrow P \equiv(13,-4) \end{aligned}$$

Distance of $$P$$ from the centre of circle $$(7,4)$$ is $$\sqrt{(13-7)^2+(-4-4)^2}=\sqrt{36+64}=10$$ units.