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(C) For the first circle $x^2 + y^2 - 4x - 6y - 3 = 0$,the center $C_1 = (2, 3)$ and radius $r_1 = \sqrt{2^2 + 3^2 - (-3)} = \sqrt{4 + 9 + 3} = 4$.For

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$3$

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(C) For the first circle $x^2 + y^2 - 4x - 6y - 3 = 0$,the center $C_1 = (2, 3)$ and radius $r_1 = \sqrt{2^2 + 3^2 - (-3)} = \sqrt{4 + 9 + 3} = 4$.For the second circle $x^2 + y^2 + 2x + 2y + 1 = 0$,the center $C_2 = (-1, -1)$ and radius $r_2 = \sqrt{(-1)^2 + (-1)^2 - 1} = \sqrt{1 + 1 - 1} = 1$.The distance between the centers $C_1C_2 = \sqrt{(2 - (-1))^2 + (3 - (-1))^2} = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = 5$.Since $C_1C_2 = r_1 + r_2$ $(5 = 4 + 1)$,the two circles touch each other externally.Therefore,the number of common tangents is $3$.

Find the number of common tangents that can be drawn to the circles $x^2 + y^2 - 4x - 6y - 3 = 0$ and $x^2 + y^2 + 2x + 2y + 1 = 0$.

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