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(D) The equation of the circle is $x^2+y^2-6x+4y+12=0$.Completing the square,we get $(x-3)^2+(y+2)^2 = 3^2+(-2)^2-12 = 9+4-12 = 1$.So,the center $O$ i

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$	an^{-1}\left(\frac{5}{12}ight)$

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(D) The equation of the circle is $x^2+y^2-6x+4y+12=0$.Completing the square,we get $(x-3)^2+(y+2)^2 = 3^2+(-2)^2-12 = 9+4-12 = 1$.So,the center $O$ is $(3,-2)$ and the radius $r$ is $1$.The distance $d$ from the point $A(2,3)$ to the center $O(3,-2)$ is $d = \sqrt{(3-2)^2+(-2-3)^2} = \sqrt{1^2+(-5)^2} = \sqrt{26}$.Let $\alpha$ be the half-angle between the tangents. In the right-angled triangle formed by the point $A$,the center $O$,and the point of tangency $P$,we have $\sin(\alpha) = \frac{r}{d} = \frac{1}{\sqrt{26}}$.Then $\cos(\alpha) = \sqrt{1-\sin^2(\alpha)} = \sqrt{1-\frac{1}{26}} = \sqrt{\frac{25}{26}} = \frac{5}{\sqrt{26}}$.Thus,$	an(\alpha) = \frac{\sin(\alpha)}{\cos(\alpha)} = \frac{1/\sqrt{26}}{5/\sqrt{26}} = \frac{1}{5}$.The total angle between the tangents is $	heta = 2\alpha$.Using the formula $	an(2\alpha) = \frac{2	an(\alpha)}{1-	an^2(\alpha)}$,we get $	an(	heta) = \frac{2(1/5)}{1-(1/5)^2} = \frac{2/5}{24/25} = \frac{2}{5} 	imes \frac{25}{24} = \frac{5}{12}$.Therefore,$	heta = 	an^{-1}\left(\frac{5}{12}ight)$.

If $\theta$ is the angle between the tangents drawn from the point $(2,3)$ to the circle $x^2+y^2-6x+4y+12=0$,then $\theta=$

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