The locus of the point of intersection of the lines $x \sin \theta +(1-\cos \theta )y=a\sin \theta$ and $x\sin \theta -(1+\cos \theta )y+a\sin \theta =0$ is

From the given equations we have
$\frac{1-\cos \theta }{\sin \theta }=\frac{a-x}{y}$ and $\frac{1+\cos \theta }{\sin \theta }=\frac{a+x}{y}$

Multiplying we get $\frac{1-{\cos }^2\theta }{{\sin }^2\theta }=\frac{a^2-x^2}{y^2} \Rightarrow x^2+y^2=a^2$