Circle And System Of Circles Question 296
Question: The equation of radical axis of the circles $ x^{2}+y^{2}+x-y+2=0 $ and $ 3x^{2}+3y^{2}-4x-12=0, $ is
[RPET 1984, 85, 86, 91, 2000]
Options:
A) $ 2x^{2}+2y^{2}-5x+y-14=0 $
B) $ 7x-3y+18=0 $
C) $ 5x-y+14=0 $
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
Radical axis is $ S_1-S_2 $
$ S_1\equiv x^{2}+y^{2}+x-y+2=0 $
$ S_2=x^{2}+y^{2}-\frac{4}{3}x-4=0 $
$ \Rightarrow S_1-S_2=\frac{7}{3}x-y+6=0 $ or $ 7x-3y+18=0 $ .
BETA
