Circle-And-System-Of-Circles Question 385
Question: The number of common tangents to the circles $ x^{2}+y^{2}=1 $ and $ x^{2}+y^{2}-4x+3=0 $ is
[DCE 2005]
Options:
1
3
2
D) None of these
Show Answer
Answer:
Correct Answer: C
Solution:
$ x^{2}+y^{2}+2x+8y-23=0 $
$ \therefore C_1(-1,-4),r_1=2\sqrt{10} $ Again $ x^{2}+y^{2}-4x-10y+9=0 $ \ $ C_2(2,5),r_2=2\sqrt{5} $ Now $ C_1C_2 $ =distance between centres.
$ \therefore $ $ C_1C_2=\sqrt{9+81}=3\sqrt{10}=9.486 $ and $ r_1+r_2=2(\sqrt{10}+\sqrt{5})=10.6 $ $ r_1-r_2=2\sqrt{5}(\sqrt{2}-1)=2\times 2.236\times 0.414=4.472\times 0.414=1.849 $ $ C_1C_2=2\sqrt{10}>r_1-r_2 $ $ r_1-r_2<C_1C_2<r_1+r_2\Rightarrow $ Two tangents can be drawn.
BETA
