The centre of the circle given by the paramet | vedclass

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Question

(B) Given the parametric equations of the circle:$x = -1 + 2\cos 	heta \Rightarrow \frac{x + 1}{2} = \cos 	heta$$y = 3 + 2\sin 	heta \Rightarrow \f

Vedclass · Accepted Answer

$(-1, 3)$

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(B) Given the parametric equations of the circle:$x = -1 + 2\cos 	heta \Rightarrow \frac{x + 1}{2} = \cos 	heta$$y = 3 + 2\sin 	heta \Rightarrow \frac{y - 3}{2} = \sin 	heta$Using the identity $\cos^2 	heta + \sin^2 	heta = 1$,we get:$\left(\frac{x + 1}{2}ight)^2 + \left(\frac{y - 3}{2}ight)^2 = 1$$(x + 1)^2 + (y - 3)^2 = 2^2$Comparing this with the standard equation of a circle $(x - h)^2 + (y - k)^2 = r^2$,where $(h, k)$ is the centre and $r$ is the radius:Here,$h = -1$ and $k = 3$.Therefore,the centre of the circle is $(-1, 3)$.

The centre of the circle given by the parametric equations $x = -1 + 2\cos \theta$ and $y = 3 + 2\sin \theta$ is:

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