If the circles x2 + y2 − 2x − 2(3 + \(\sqrt{7}\))y + 8 + 6\(\sqrt{7}\) = 0 and x2 + y2 − 8x − 6y + k2 = 0, k ∈ ℤ, have exactly two common tangents, then the number of possible values for k is

1
8
2
5
3
9
4
11

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