A) \[1+\sqrt{1+{{m}^{2}}}\]
B) \[1-\sqrt{{{m}^{2}}+1}\]
C) \[2(1+\sqrt{1+{{m}^{2}}})\]
D) \[2+\sqrt{1+{{m}^{2}}}\]
Correct Answer: C
Solution :
Apply for tangency of line, centre being (0, 2) and radius = 2 \[\left| \frac{-2+c}{\sqrt{1+{{m}^{2}}}} \right|=2\Rightarrow {{c}^{2}}-4c+4=4+4{{m}^{2}}\] \[\Rightarrow c=\frac{4\pm \sqrt{16+16{{m}^{2}}}}{2}\] or\[c=2\pm 2\sqrt{1+{{m}^{2}}}\] Most correct answer is\[c=2(1+\sqrt{1+{{m}^{2}}})\].
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