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

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(B) Given the equation of the circle: $x^2+y^2-6x-2y+9=0$. Completing the square for $x$ and $y$ terms: $(x^2-6x+9) + (y^2-2y+1) = -9+9+1$. $(x-3)^2 +

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

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(B) Given the equation of the circle: $x^2+y^2-6x-2y+9=0$. Completing the square for $x$ and $y$ terms: $(x^2-6x+9) + (y^2-2y+1) = -9+9+1$. $(x-3)^2 + (y-1)^2 = 1^2$. The standard form of a circle is $(x-h)^2 + (y-k)^2 = r^2$,where the center is $(h, k) = (3, 1)$ and the radius $r = 1$. The parametric equations are given by $x = h + r \cos 	heta$ and $y = k + r \sin 	heta$. Substituting the values: $x = 3 + 1 \cos 	heta$ and $y = 1 + 1 \sin 	heta$. Thus,$x = 3 + \cos 	heta$ and $y = 1 + \sin 	heta$.

The parametric equations of the circle $x^2+y^2-6x-2y+9=0$ are

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