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

Solution :
$$\eqalign{ & \frac{{dx}}{{d\theta }} = a\left( { - \sin \theta + \sin \theta + \theta \cos \theta } \right) = a\theta \cos \theta \,......\left( 1 \right) \cr & \frac{{dy}}{{d\theta }} = a\left( {\cos \theta - \cos \theta + \theta \sin \theta } \right) = a\theta \sin \theta \,......\left( 2 \right) \cr} $$
Dividing (2) by (1), we get
$$\eqalign{ & \frac{{dy}}{{dx}} = \tan \theta \,\left( {{\text{slope of tangent}}} \right) \cr & \therefore {\text{Slope of normal }} = - \cot \theta \cr & \therefore {\text{Equation of normal is}} \cr & y - a\left( {\sin \theta - \theta \cos \theta } \right) = - \frac{{\cos \theta }}{{\sin \theta }}\left( {x - a\left( {\cos \theta + \theta \sin \theta } \right)} \right) \cr & \Rightarrow y\sin \theta - a{\sin ^2}\theta + a\sin \theta \cos \theta \cr & = - x\cos \theta + a{\cos ^2}\theta + a\theta \sin \theta \cos \theta \cr & \Rightarrow x\cos \theta + y\sin \theta = a \cr} $$
As $$\theta $$ varies inclination is not constant.
∴ (A) is not correct.
Clearly does not pass through (0, 0).
It's distance from origin $$ = \left| {\frac{a}{{\sqrt {{{\cos }^2}\theta + {{\sin }^2}\theta } }}} \right| = a$$
which is constant