The normal to the curve $x=a(1+\cos \theta ),y=a\sin \theta$ at '$\theta$ ' always passes through the fixed point

The curve equations are $x=a+a\cos \theta$ and $y=a\sin \theta .$
Differentiating with respect to $\theta$:
$\frac{dx}{d\theta }=-a\sin \theta$ and $\frac{dy}{d\theta }=a\cos \theta .$
Slope of tangent $=\frac{dy/d\theta }{dx/d\theta }=\frac{a\cos \theta }{-a\sin \theta }=-\cot \theta .$
Slope of normal $=\frac{-1}{\text{Slope of tangent}}=\frac{-1}{-\cot \theta }=\tan \theta .$
Equation of normal at point $(a(1+\cos \theta ),a\sin \theta )$:
$y-a\sin \theta =\tan \theta [x-a(1+\cos \theta )]$
$y-a\sin \theta =\frac{\sin \theta }{\cos \theta }[x-a-a\cos \theta ]$
$y\cos \theta -a\sin \theta \cos \theta =x\sin \theta -a\sin \theta -a\sin \theta \cos \theta$
$y\cos \theta =x\sin \theta -a\sin \theta \Rightarrow y\cos \theta =\sin \theta (x-a).$
Substituting the point $(a,0)$:
$0\cdot \cos \theta =\sin \theta (a-a)\Rightarrow 0=0.$
The equation is satisfied for $(a,0)$ regardless of $\theta .$
Alternatively, eliminating $\theta$: $(x-a{)}^2+y^2=a^2({\cos }^2\theta +{\sin }^2\theta )=a^2$.
This is a circle with center $(a,0)$ and radius $a.$ The normal to any point on a circle always passes through its center $(a,0)$.