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(B) To find the locus,we eliminate the parameter $	heta$.Given equations are:$x = a(1 - \cos 	heta )$ $\Rightarrow x - a = -a \cos 	heta$ $\Rightar

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$A$ circle

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(B) To find the locus,we eliminate the parameter $	heta$.Given equations are:$x = a(1 - \cos 	heta )$ $\Rightarrow x - a = -a \cos 	heta$ $\Rightarrow \cos 	heta = -\frac{x - a}{a} = \frac{a - x}{a}$ .....$(i)$$y = a \sin 	heta \Rightarrow \sin 	heta = \frac{y}{a}$ .....$(ii)$Using the identity $\sin^2 	heta + \cos^2 	heta = 1$,we substitute $(i)$ and $(ii)$:$(\frac{y}{a})^2 + (\frac{a - x}{a})^2 = 1$$\frac{y^2}{a^2} + \frac{a^2 - 2ax + x^2}{a^2} = 1$$y^2 + a^2 - 2ax + x^2 = a^2$$x^2 + y^2 - 2ax = 0$This is the equation of a circle with center $(a, 0)$ and radius $a$.

If the coordinates of a point are given by the equations $x = a(1 - \cos \theta )$ and $y = a\sin \theta $,then the locus of the point will be

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