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


A
2
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B
8
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C
9
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D
5
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Solution

The correct option is D $$9$$
$$S_{1}:x^{2}+y^{2}=9$$
$$C_{1}\equiv (0,0)$$ and $$r_{1}=3$$
$$S_{2}:x^{2}+y^{2}-8x-6y+n^{2}=0$$
$$C_{2}\equiv (4,3)$$ and $$r_{2}=\sqrt{25-n^{2}}$$
 $$r_{2}=\sqrt{25-n^{2}}\gt0$$  so    $$-5\leq n\leq 5$$ 
$$\therefore $$ Given $$S_{1}$$ and $$S_{2}$$ have two common tangents, $$S_{1}$$ and $$S_{2}$$ are intersecting each other.
$$\therefore r_{1}+r_{2}>C_{1}C_{2} > |r_1-r_2|$$
$$C_1C_2 > |r_1-r_2|$$ is true for all $$n$$.

$$3+\sqrt{25-n^{2}}>5$$
$$\sqrt{25-n^{2}}>2$$
$$25-n^{2}>4$$
$$n^{2}<21$$
$$(n-\sqrt{21})(n+\sqrt{21})<0$$
$$n\in (-\sqrt{21},\sqrt{21})$$    ...(1) 
and $$-5\leq n\leq 5$$    ...(2)

From (1) & (2) $$n\in (-\sqrt{21},\sqrt{21}),n\in Z.$$
$$n=-4,-3,-2,-1,0,1,2,3,4$$
$$\Rightarrow $$ Number of values of $$n$$ are $$9$$

Mathematics

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