The coordinates of a point on the circle (x - | vedclass

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(B) The given equation of the circle is $(x - 1)^2 + (y - 1)^2 = 4$.This is in the form $(x - h)^2 + (y - k)^2 = r^2$,where the center $(h, k) = (1, 1

Vedclass · Accepted Answer

$(1 + 2 \cos \alpha, 1 + 2 \sin \alpha)$

Vedclass · Answer

(B) The given equation of the circle is $(x - 1)^2 + (y - 1)^2 = 4$.This is in the form $(x - h)^2 + (y - k)^2 = r^2$,where the center $(h, k) = (1, 1)$ and the radius $r = \sqrt{4} = 2$.The parametric coordinates of a point on a circle with center $(h, k)$ and radius $r$ are given by $(h + r \cos \alpha, k + r \sin \alpha)$.Substituting the values,we get $(1 + 2 \cos \alpha, 1 + 2 \sin \alpha)$.Thus,the correct option is $B$.

The coordinates of a point on the circle $(x - 1)^2 + (y - 1)^2 = 4$ are given by:

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