Circle And System Of Circles Question 26
Question: The number of common tangents to two circles $ x^{2}+y^{2}=4 $ and $ x^{2}-y^{2}-8x+12=0 $ is
[EAMCET 1990]
Options:
A) 1
B) 2
C) 3
D) 4
Show Answer
Answer:
Correct Answer: C
Solution:
Here $ C_1=(0,\ 0),\ r_1=2,\ C_2=(4,\ 0) $ , $ r_2=2 $ Here $ C_1C_2=4=r_1+r_2 $ . Thus two circles touch externally,
Hence, the number of common tangents is 3.
BETA
