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(D) Given circles are:$S_1: x^2 + y^2 - 2y - 3 = 0$Centre $C_1 = (0, 1)$,Radius $r_1 = \sqrt{0^2 + (-1)^2 - (-3)} = \sqrt{1 + 3} = 2$.$S_2: x^2 + y^2

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$(2, 5)$

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(D) Given circles are:$S_1: x^2 + y^2 - 2y - 3 = 0$Centre $C_1 = (0, 1)$,Radius $r_1 = \sqrt{0^2 + (-1)^2 - (-3)} = \sqrt{1 + 3} = 2$.$S_2: x^2 + y^2 - 8x - 18y + 93 = 0$Centre $C_2 = (4, 9)$,Radius $r_2 = \sqrt{(-4)^2 + (-9)^2 - 93} = \sqrt{16 + 81 - 93} = \sqrt{4} = 2$.Since the radii of both circles are equal $(r_1 = r_2 = 2)$,the smallest circle touching both circles externally will have its diameter equal to the distance between the two circles along the line joining their centres.The centre of this smallest circle is the midpoint of the line segment joining the centres $C_1$ and $C_2$.Midpoint $= \left( \frac{0 + 4}{2}, \frac{1 + 9}{2} \right) = (2, 5)$.

The centre of the smallest circle touching the circles $x^2 + y^2 - 2y - 3 = 0$ and $x^2 + y^2 - 8x - 18y + 93 = 0$ is:

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