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(D) The given equation of the circle is $x^{2}+y^{2}+4x-6y+9 \sin^{2} \alpha + 13 \cos^{2} \alpha = 0$. Comparing this with the general equation $x^{2

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

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(D) The given equation of the circle is $x^{2}+y^{2}+4x-6y+9 \sin^{2} \alpha + 13 \cos^{2} \alpha = 0$. Comparing this with the general equation $x^{2}+y^{2}+2gx+2fy+c=0$,we get the center $C = (-2, 3)$. The radius $r$ is given by $\sqrt{g^{2}+f^{2}-c} = \sqrt{(-2)^{2}+(3)^{2}-(9 \sin^{2} \alpha + 13 \cos^{2} \alpha)}$. $r = \sqrt{4+9-9 \sin^{2} \alpha - 13 \cos^{2} \alpha} = \sqrt{13-9 \sin^{2} \alpha - 13(1-\sin^{2} \alpha)} = \sqrt{13-9 \sin^{2} \alpha - 13 + 13 \sin^{2} \alpha} = \sqrt{4 \sin^{2} \alpha} = 2 \sin \alpha$. Let $P(h, k)$ be a point on the locus. The angle between the tangents is $2 \alpha$,so the angle between the line $PC$ and a tangent is $\alpha$. In the right-angled triangle $	riangle PAC$,$\sin \alpha = \frac{AC}{PC} = \frac{r}{PC}$. Thus,$PC = \frac{r}{\sin \alpha} = \frac{2 \sin \alpha}{\sin \alpha} = 2$. $PC^{2} = 4$. $(h+2)^{2} + (k-3)^{2} = 4$. $h^{2}+4h+4 + k^{2}-6k+9 = 4$. $h^{2}+k^{2}+4h-6k+9 = 0$. Replacing $(h, k)$ with $(x, y)$,the locus is $x^{2}+y^{2}+4x-6y+9=0$.

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 $2 \alpha$. The equation of the locus of the point $P$ is

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