The normal to the curve $x=a(\cos \theta +\theta \sin \theta )$, $y=a(\sin \theta -\theta \cos \theta )$ at any point $\theta$ is such that:

First, find the slope of the tangent $\frac{dy}{dx}$:$\frac{dx}{d\theta }=a(-\sin \theta +\sin \theta +\theta \cos \theta )=a\theta \cos \theta$$\frac{dy}{d\theta }=a(\cos \theta -(\cos \theta -\theta \sin \theta ))=a\theta \sin \theta$$\frac{dy}{dx}=\frac{dy/d\theta }{dx/d\theta }=\frac{a\theta \sin \theta }{a\theta \cos \theta }=\tan \theta$.The slope of the normal is $m_n=-\frac{1}{\tan \theta }=-\cot \theta =\frac{-\cos \theta }{\sin \theta }$.Equation of the normal:$y-a(\sin \theta -\theta \cos \theta )=-\frac{\cos \theta }{\sin \theta }[x-a(\cos \theta +\theta \sin \theta )]$$y\sin \theta -a{\sin }^2\theta +a\theta \sin \theta \cos \theta =-x\cos \theta +a{\cos }^2\theta +a\theta \sin \theta \cos \theta$$x\cos \theta +y\sin \theta =a({\cos }^2\theta +{\sin }^2\theta )$$x\cos \theta +y\sin \theta =a$ The perpendicular distance from the origin $(0,0)$ to this line is:$d=\frac{|0\cdot \cos \theta +0\cdot \sin \theta -a|}{\sqrt{{\cos }^2\theta +{\sin }^2\theta }}=\frac{|-a|}{1}=a$.Since $d$ is independent of $\theta$, the normal is at a constant distance $a$ from the origin.