If T_1 T_1^{prime} and T_2 T_2^{prime} are th | vedclass

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(D) For the circle $S = x^2 + y^2 - 2x - 4y - 4 = 0$, the center $C_1 = (1, 2)$ and radius $r_1 = \sqrt{1^2 + 2^2 - (-4)} = \sqrt{9} = 3$.For the circ

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$2$

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(D) For the circle $S = x^2 + y^2 - 2x - 4y - 4 = 0$, the center $C_1 = (1, 2)$ and radius $r_1 = \sqrt{1^2 + 2^2 - (-4)} = \sqrt{9} = 3$.For the circle $S^{\prime} = x^2 + y^2 + 4x + 4y + 4 = 0$, the center $C_2 = (-2, -2)$ and radius $r_2 = \sqrt{(-2)^2 + (-2)^2 - 4} = \sqrt{4} = 2$.The distance between the centers $C_1 C_2 = \sqrt{(1 - (-2))^2 + (2 - (-2))^2} = \sqrt{3^2 + 4^2} = \sqrt{25} = 5$.The length of the direct common tangent is given by $L = \sqrt{(C_1 C_2)^2 - (r_1 - r_2)^2}$.$L = \sqrt{5^2 - (3 - 2)^2} = \sqrt{25 - 1} = \sqrt{24} = 2 \sqrt{6}$.Since $T_1 T_1^{\prime}$ represents the length of the common tangent, the distance is $2 \sqrt{6}$.

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