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(D) The given circle is $x^2+y^2+4x-6y+9 \sin^2 \alpha+13 \cos^2 \alpha=0$. Comparing with $x^2+y^2+2gx+2fy+c=0$,we have $g=2$,$f=-3$,and $c=9 \sin^2

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$x^2+y^2+4x-6y+9=0$

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(D) The given circle is $x^2+y^2+4x-6y+9 \sin^2 \alpha+13 \cos^2 \alpha=0$. Comparing with $x^2+y^2+2gx+2fy+c=0$,we have $g=2$,$f=-3$,and $c=9 \sin^2 \alpha+13 \cos^2 \alpha$. The center $C$ is $(-g, -f) = (-2, 3)$. The radius $r$ is $\sqrt{g^2+f^2-c} = \sqrt{4+9-(9 \sin^2 \alpha+13 \cos^2 \alpha)} = \sqrt{13-9 \sin^2 \alpha-13 \cos^2 \alpha} = \sqrt{13(1-\cos^2 \alpha)-9 \sin^2 \alpha} = \sqrt{13 \sin^2 \alpha-9 \sin^2 \alpha} = \sqrt{4 \sin^2 \alpha} = 2 \sin \alpha$. Let $P(x_1, y_1)$ be the point. The distance $PC = \sqrt{(x_1+2)^2+(y_1-3)^2} = \sqrt{x_1^2+y_1^2+4x_1-6y_1+13}$. In the right-angled triangle $	riangle PAC$,$\sin \alpha = \frac{AC}{PC} = \frac{r}{PC}$. Thus,$\sin \alpha = \frac{2 \sin \alpha}{\sqrt{x_1^2+y_1^2+4x_1-6y_1+13}}$. Squaring both sides,$\sin^2 \alpha = \frac{4 \sin^2 \alpha}{x_1^2+y_1^2+4x_1-6y_1+13}$. $x_1^2+y_1^2+4x_1-6y_1+13 = 4$. $x_1^2+y_1^2+4x_1-6y_1+9 = 0$. Replacing $(x_1, y_1)$ with $(x, y)$,the locus is $x^2+y^2+4x-6y+9=0$.

If the angle between a pair of tangents drawn from a point $P$ to the circle $x^2+y^2+4x-6y+9 \sin^2 \alpha+13 \cos^2 \alpha=0$ is $2 \alpha$,then the equation of the locus of $P$ is

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