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(B) Given equations of the straight lines are:$x \sin 	heta + (1 - \cos 	heta) y = a \sin 	heta$ ... $(i)$$x \sin 	heta - (1 + \cos 	heta) y = -a

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(B) Given equations of the straight lines are:$x \sin 	heta + (1 - \cos 	heta) y = a \sin 	heta$ ... $(i)$$x \sin 	heta - (1 + \cos 	heta) y = -a \sin 	heta$ ... $(ii)$Subtracting Eq. $(ii)$ from Eq. $(i)$,we get:$(1 - \cos 	heta) y - (-(1 + \cos 	heta) y) = a \sin 	heta - (-a \sin 	heta)$$y(1 - \cos 	heta + 1 + \cos 	heta) = 2a \sin 	heta$$2y = 2a \sin 	heta$$y = a \sin 	heta$Substituting the value of $y$ in Eq. $(i)$:$x \sin 	heta + (1 - \cos 	heta)(a \sin 	heta) = a \sin 	heta$Dividing by $\sin 	heta$ (assuming $\sin 	heta 
eq 0$):$x + a - a \cos 	heta = a$$x = a \cos 	heta$Now,to find the locus,we square and add $x$ and $y$:$x^2 + y^2 = (a \cos 	heta)^2 + (a \sin 	heta)^2$$x^2 + y^2 = a^2(\cos^2 	heta + \sin^2 	heta)$$x^2 + y^2 = a^2$This represents a circle with center $(0, 0)$ and radius $a$.

For any value of $\theta$,if the straight lines $x \sin \theta + (1 - \cos \theta) y = a \sin \theta$ and $x \sin \theta - (1 + \cos \theta) y + a \sin \theta = 0$ intersect at $P(\theta)$,then the locus of $P(\theta)$ is a

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