Question

# 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
B
8
C
9
D
5

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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