If theta is a parameter,then the parametric e | vedclass

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(B) The given equation of the circle is $x^{2}+y^{2}-6x+4y-3=0$. Completing the square,we get: $(x^{2}-6x+9) + (y^{2}+4y+4) = 3+9+4$ $(x-3)^{2} + (y+2

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$x=3+4 \cos 	heta$ and $y=-2+4 \sin 	heta$

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(B) The given equation of the circle is $x^{2}+y^{2}-6x+4y-3=0$. Completing the square,we get: $(x^{2}-6x+9) + (y^{2}+4y+4) = 3+9+4$ $(x-3)^{2} + (y+2)^{2} = 16$ $(x-3)^{2} + (y+2)^{2} = 4^{2}$ Comparing this with the standard form $(x-h)^{2} + (y-k)^{2} = r^{2}$,we identify the center $(h, k) = (3, -2)$ and radius $r = 4$. The parametric equations of a circle are given by $x = h + r \cos 	heta$ and $y = k + r \sin 	heta$. Substituting the values,we get $x = 3 + 4 \cos 	heta$ and $y = -2 + 4 \sin 	heta$.

If $\theta$ is a parameter,then the parametric equations of the circle $x^{2}+y^{2}-6x+4y-3=0$ are given by

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