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(D) The equation of the 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 =

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

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(D) The equation of the 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 $g = 2, f = -3, c = 9{\sin ^2}\alpha + 13{\cos ^2}\alpha$.The centre is $C(-g, -f) = (-2, 3)$.The radius $r = \sqrt{g^2 + f^2 - c} = \sqrt{4 + 9 - (9{\sin ^2}\alpha + 13{\cos ^2}\alpha)} = \sqrt{13 - 9{\sin ^2}\alpha - 13{\cos ^2}\alpha} = \sqrt{13(1 - {\cos ^2}\alpha) - 9{\sin ^2}\alpha} = \sqrt{13{\sin ^2}\alpha - 9{\sin ^2}\alpha} = \sqrt{4{\sin ^2}\alpha} = 2\sin \alpha$.Let $P(h, k)$ be a point on the locus. The angle between the tangents is $2\alpha$,so $\angle APC = \alpha$.In the right-angled triangle $	riangle PAC$,$\sin \alpha = \frac{AC}{PC} = \frac{r}{PC} = \frac{2\sin \alpha}{\sqrt{(h + 2)^2 + (k - 3)^2}}$.Thus,$\sqrt{(h + 2)^2 + (k - 3)^2} = 2$.Squaring both sides,$(h + 2)^2 + (k - 3)^2 = 4$.Expanding,$h^2 + 4h + 4 + k^2 - 6k + 9 = 4$,which simplifies to $h^2 + k^2 + 4h - 6k + 9 = 0$.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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