If P(frac{pi}{3}) and Q(frac{2pi}{3}) represe | vedclass

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(B) The given equation of the circle is $x^2+y^2-4x+6y-12=0$. Completing the square,we get $(x-2)^2+(y+3)^2 = 12+4+9 = 25 = 5^2$. Thus,the center is $

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$5$

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(B) The given equation of the circle is $x^2+y^2-4x+6y-12=0$. Completing the square,we get $(x-2)^2+(y+3)^2 = 12+4+9 = 25 = 5^2$. Thus,the center is $(2, -3)$ and the radius $r = 5$. The parametric coordinates are given by $x = 2 + 5\cos	heta$ and $y = -3 + 5\sin	heta$. For point $P$ with $	heta = \frac{\pi}{3}$,$P = (2 + 5\cos\frac{\pi}{3}, -3 + 5\sin\frac{\pi}{3}) = (2 + \frac{5}{2}, -3 + \frac{5\sqrt{3}}{2}) = (\frac{9}{2}, -3 + \frac{5\sqrt{3}}{2})$. For point $Q$ with $	heta = \frac{2\pi}{3}$,$Q = (2 + 5\cos\frac{2\pi}{3}, -3 + 5\sin\frac{2\pi}{3}) = (2 - \frac{5}{2}, -3 + \frac{5\sqrt{3}}{2}) = (-\frac{1}{2}, -3 + \frac{5\sqrt{3}}{2})$. The distance $PQ$ is $\sqrt{(\frac{9}{2} - (-\frac{1}{2}))^2 + (-3 + \frac{5\sqrt{3}}{2} - (-3 + \frac{5\sqrt{3}}{2}))^2} = \sqrt{(\frac{10}{2})^2 + 0^2} = 5$.

If $P(\frac{\pi}{3})$ and $Q(\frac{2\pi}{3})$ represent two points on the circle $x^2+y^2-4x+6y-12=0$ in parametric form,then the length of the chord $PQ$ is

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