If x = 2 + 3 cos theta and y = 1 - 3 sin thet | vedclass

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(B) Given that,$x = 2 + 3 \cos 	heta \implies x - 2 = 3 \cos 	heta \implies \cos 	heta = \frac{x - 2}{3}$ $y = 1 - 3 \sin 	heta \implies y - 1 = -

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$(2, 1), 3$

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(B) Given that,$x = 2 + 3 \cos 	heta \implies x - 2 = 3 \cos 	heta \implies \cos 	heta = \frac{x - 2}{3}$ $y = 1 - 3 \sin 	heta \implies y - 1 = -3 \sin 	heta \implies \sin 	heta = \frac{y - 1}{-3}$ We know the identity $\sin^2 	heta + \cos^2 	heta = 1$. Substituting the values,we get: $\left(\frac{y - 1}{-3}ight)^2 + \left(\frac{x - 2}{3}ight)^2 = 1$ $\frac{(y - 1)^2}{9} + \frac{(x - 2)^2}{9} = 1$ $(x - 2)^2 + (y - 1)^2 = 9$ Comparing this with the standard equation of a circle $(x - h)^2 + (y - k)^2 = r^2$,we get the centre $(h, k) = (2, 1)$ and radius $r = \sqrt{9} = 3$.

If $x = 2 + 3 \cos \theta$ and $y = 1 - 3 \sin \theta$ represent a circle,then the centre and radius are:

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