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  • There are two circles whose equations are and <img alt="\mathrm{x^2+y^

There are two circles whose equations are \mathrm{x^2+y^2=9} and \mathrm{x^2+y^2-8 x-6 y+n^2=0, n \in Z}. If the two circles have exactly two common tangents, then the number of possible values of n is

Option: 1

2


Option: 2

8


Option: 3

9


Option: 4

None of these


Answers (1)

best_answer

\mathrm{\text { For } x^2+y^2=9 \text {, the centre }=(0,0) \text { and the radius }=3}

\mathrm{\text { For } x^2+y^2-8 x-6 y+n^2=0 \text {. The centre }=(4,3) \text { and the radius }=\sqrt{(4)^2+(3)^2-n^2}}

\mathrm{\therefore \quad 4^2+3^2-n^2>0 \text { or } n^2<5^2 \text { or }-5<n<5 \text {. }}

Circles should cut to have exactly two common tangents.

\mathrm{\text { So, } r_1+r_2>C_1 C_2, \therefore 3+\sqrt{25-n^2}>\sqrt{(4)^2+(3)^2} \text { or } \sqrt{25-n^2}>2 \text { or } 25-n^2>4}

\mathrm{\therefore n^2<21 \text { or }-\sqrt{21}<n<\sqrt{21}}

Therefore, common values of n should satisfy \mathrm{ -\sqrt{21}<n<\sqrt{21}.}

But \mathrm{ n \in Z, } So,\mathrm{ n=-4,-3, \ldots \ldots . .3,4 . \therefore}  Number of possible values of \mathrm{n=9}.

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Rakesh

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