The equation of the locus of the point of int | vedclass

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(D) Given equations are: $x \sin 	heta + (1 - \cos 	heta) y = a \sin 	heta$  $(1)$ $x \sin 	heta - (1 + \cos 	heta) y = -a \sin 	heta$  $(2)$ Su

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$x^2 + y^2 = a^2$

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(D) Given equations are: $x \sin 	heta + (1 - \cos 	heta) y = a \sin 	heta$  $(1)$ $x \sin 	heta - (1 + \cos 	heta) y = -a \sin 	heta$  $(2)$ Subtracting $(2)$ from $(1)$: $(1 - \cos 	heta + 1 + \cos 	heta) y = a \sin 	heta + a \sin 	heta$ $2y = 2a \sin 	heta \implies y = a \sin 	heta$ Adding $(1)$ and $(2)$: $2x \sin 	heta + (1 - \cos 	heta - 1 - \cos 	heta) y = 0$ $2x \sin 	heta - 2y \cos 	heta = 0$ $x \sin 	heta = y \cos 	heta$ Substitute $y = a \sin 	heta$: $x \sin 	heta = (a \sin 	heta) \cos 	heta$ $x = a \cos 	heta$ Now,$x^2 + y^2 = (a \cos 	heta)^2 + (a \sin 	heta)^2 = a^2(\cos^2 	heta + \sin^2 	heta) = a^2$ Thus,the locus is $x^2 + y^2 = a^2$.

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

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