If the circles x^2+y^2-2x-2(3+sqrt{7})y+8+6sq | vedclass

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(C) The centers of the given circles are $C_1(1, 3+\sqrt{7})$ and $C_2(4, 3)$.Their radii are $r_1 = \sqrt{1^2 + (3+\sqrt{7})^2 - (8+6\sqrt{7})} = \sq

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$9$

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(C) The centers of the given circles are $C_1(1, 3+\sqrt{7})$ and $C_2(4, 3)$.Their radii are $r_1 = \sqrt{1^2 + (3+\sqrt{7})^2 - (8+6\sqrt{7})} = \sqrt{1 + 9 + 7 + 6\sqrt{7} - 8 - 6\sqrt{7}} = \sqrt{9} = 3$.And $r_2 = \sqrt{4^2 + 3^2 - k^2} = \sqrt{25-k^2}$.The distance between the centers is $C_1C_2 = \sqrt{(4-1)^2 + (3-(3+\sqrt{7}))^2} = \sqrt{3^2 + (-\sqrt{7})^2} = \sqrt{9+7} = 4$.For the circles to have exactly two common tangents,they must intersect at two distinct points,which implies $|r_1 - r_2| First,$C_1C_2 Second,$|r_1 - r_2| This implies $-4 Since $\sqrt{25-k^2} \ge 0$,we have $0 \le \sqrt{25-k^2}  -24$.Combining $k^2 Since $k \in \mathbb{Z}$,$k^2 \in \{0, 1, 4, 9, 16\}$.Possible values for $k$ are $0, \pm 1, \pm 2, \pm 3, \pm 4$.Total number of values is $1 + 4 + 4 = 9$.

If the circles $x^2+y^2-2x-2(3+\sqrt{7})y+8+6\sqrt{7}=0$ and $x^2+y^2-8x-6y+k^2=0, k \in \mathbb{Z}$,have exactly two common tangents,then the number of possible values of $k$ is

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