The parametric equations of the circle x^2+y^ | vedclass

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(A) The given equation of the circle is $x^2+y^2+2x-4y-4=0$. Completing the square for $x$ and $y$ terms: $(x^2+2x+1) + (y^2-4y+4) - 4 - 1 - 4 = 0$ $(

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$x=-1+3 \cos 	heta, y=2+3 \sin 	heta$

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(A) The given equation of the circle is $x^2+y^2+2x-4y-4=0$. Completing the square for $x$ and $y$ terms: $(x^2+2x+1) + (y^2-4y+4) - 4 - 1 - 4 = 0$ $(x+1)^2 + (y-2)^2 = 9$ $(x+1)^2 + (y-2)^2 = 3^2$ Comparing this with the standard form $(x-h)^2 + (y-k)^2 = r^2$,we get the center $(h, k) = (-1, 2)$ and radius $r = 3$. The parametric equations are given by $x = h + r \cos 	heta$ and $y = k + r \sin 	heta$. Substituting the values,we get $x = -1 + 3 \cos 	heta$ and $y = 2 + 3 \sin 	heta$.

The parametric equations of the circle $x^2+y^2+2x-4y-4=0$ are

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