Circle And System Of Circles Question 265
Question: The equation of radical axis of the circles $ 2x^{2}+2y^{2}-7x=0 $ and $ x^{2}+y^{2}-4y-7=0 $ is
[RPET 1996]
Options:
A) $ 7x+8y+14=0 $
B) $ 7x-8y+14=0 $
C) $ 7x-8y-14=0 $
D) None of these
Show Answer
Answer:
Correct Answer: C
Solution:
Equation of radical axis, $ S_1-S_2=0 $ i.e., $ (2x^{2}+2y^{2}-7x)-(2x^{2}+2y^{2}-8y-14)=0 $
$ \Rightarrow -7x+8y+14=0 $ ,
$ \therefore \ 7x-8y-14=0 $ .
BETA
