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  • If the circles <img alt="\mathrm{x^{2}+y^{2}-8 x+2 y+8=0\, and \, x^{2}+y^{2}-2 x-6 y-a^{2}=0}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7Bx%5E%7B2%7D&plus;y%5E%7B2%7D-8%2

If the circles \mathrm{x^{2}+y^{2}-8 x+2 y+8=0\, and \, x^{2}+y^{2}-2 x-6 y-a^{2}=0} have exactly two common tangents,

Option: 1

1<|a|<8


Option: 2

2<|a|<8


Option: 3

3<|a|<8


Option: 4

\quad 4<|a|<8


Answers (1)

best_answer

Centres of the circles are (4,-1) and (1,3). Their radii are 3 and |a| respectively. They will have exactly two common tangents if they meet in two distinct points. That means

\begin{aligned} & |3-| a||<\sqrt{3^{2}+4^{2}}<3+|a| \\ & \Rightarrow|3-| a||<5<3+|a| \end{aligned}

From \mathrm{|3-| a||>5} we get \mathrm{a \in(-\infty,-2) \cup(2, \infty)}

 From \mathrm{|3+| a||<5} we get  \mathrm{a \in(-8,8)}

\mathrm{a \in(-8,-2) \cup(2,8) i.e. 2<|a|<8}

Hence (B) is the correct answer.

Posted by

Gunjita

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