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(C) The given equation of the circle is $x^2+y^2-6x+4y-12=0$.Completing the square,we get $(x-3)^2+(y+2)^2=5^2$.The center is $(h, k) = (3, -2)$ and t

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$x+\sqrt{3}y-3(1+\sqrt{3})=0$

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(C) The given equation of the circle is $x^2+y^2-6x+4y-12=0$.Completing the square,we get $(x-3)^2+(y+2)^2=5^2$.The center is $(h, k) = (3, -2)$ and the radius $r=5$.The parametric coordinates are $x=3+5\cos	heta$ and $y=-2+5\sin	heta$.For point $A$ with $	heta=30^{\circ}$,$A = (3+5\cos 30^{\circ}, -2+5\sin 30^{\circ}) = (3+\frac{5\sqrt{3}}{2}, -2+\frac{5}{2}) = (\frac{6+5\sqrt{3}}{2}, \frac{1}{2})$.For point $B$ with $	heta=90^{\circ}$,$B = (3+5\cos 90^{\circ}, -2+5\sin 90^{\circ}) = (3, 3)$.The slope of chord $AB$ is $m = \frac{3-1/2}{3-(6+5\sqrt{3})/2} = \frac{5/2}{(6-5\sqrt{3})/2} = \frac{5}{6-5\sqrt{3}}$.Wait,calculating slope $m = \frac{3-0.5}{3-(3+2.5\sqrt{3})} = \frac{2.5}{-2.5\sqrt{3}} = -\frac{1}{\sqrt{3}}$.Using the point-slope form $y-y_1 = m(x-x_1)$ at point $B(3, 3)$:$y-3 = -\frac{1}{\sqrt{3}}(x-3)$.$\sqrt{3}y-3\sqrt{3} = -x+3$.$x+\sqrt{3}y-3-3\sqrt{3} = 0$.$x+\sqrt{3}y-3(1+\sqrt{3}) = 0$.

If the parametric values of two points $A$ and $B$ on the circle $x^2+y^2-6x+4y-12=0$ are $30^{\circ}$ and $90^{\circ}$ respectively,then the equation of chord $AB$ is

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