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(D) The given circle is $x^{2}+y^{2}-4x-6y-3=0$. Comparing this with the general equation $x^{2}+y^{2}+2gx+2fy+c=0$, we get the centre $O(2, 3)$ and r

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$3(x^{2}+y^{2})-12x-18y-25=0$

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(D) The given circle is $x^{2}+y^{2}-4x-6y-3=0$. Comparing this with the general equation $x^{2}+y^{2}+2gx+2fy+c=0$, we get the centre $O(2, 3)$ and radius $r = \sqrt{g^{2}+f^{2}-c} = \sqrt{2^{2}+3^{2}-(-3)} = \sqrt{4+9+3} = \sqrt{16} = 4$.Let $R(h, k)$ be the point of intersection of the tangents $PQ$ and $MN$. The line $OR$ bisects $\angle AOB$. Thus, $\angle AOR = \angle BOR = \frac{1}{2} \angle AOB = \frac{1}{2} (\pi/3) = \pi/6$ (or $30^{\circ}$).In the right-angled triangle $	riangle OAR$, we have $\cos(\angle AOR) = \frac{OA}{OR}$.$\cos(30^{\circ}) = \frac{r}{OR} \implies \frac{\sqrt{3}}{2} = \frac{4}{OR} \implies OR = \frac{8}{\sqrt{3}}$.Squaring both sides, $OR^{2} = \frac{64}{3}$.Since $O$ is $(2, 3)$ and $R$ is $(h, k)$, $OR^{2} = (h-2)^{2} + (k-3)^{2}$.So, $(h-2)^{2} + (k-3)^{2} = \frac{64}{3}$.$3(h^{2}-4h+4 + k^{2}-6k+9) = 64$.$3(h^{2}+k^{2}-4h-6k+13) = 64$.$3(h^{2}+k^{2}) - 12h - 18k + 39 = 64$.$3(h^{2}+k^{2}) - 12h - 18k - 25 = 0$.Replacing $(h, k)$ with $(x, y)$, the locus is $3(x^{2}+y^{2}) - 12x - 18y - 25 = 0$.

Let $PQ$ and $MN$ be two straight lines touching the circle $x^{2}+y^{2}-4x-6y-3=0$ at the points $A$ and $B$ respectively. Let $O$ be the centre of the circle and $\angle AOB=\pi/3$. Then the locus of the point of intersection of the lines $PQ$ and $MN$ is:

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