Circle And System Of Circles Question 251
Question: Polar of origin (0, 0) with respect to the circle $ x^{2}+y^{2}+2\lambda x+2\mu y+c=0 $ touches the circle $ x^{2}+y^{2}=r^{2} $ , if
[RPET 1992]
Options:
A) $ c=r({{\lambda }^{2}}+{{\mu }^{2}}) $
B) $ r=c,({{\lambda }^{2}}+{{\mu }^{2}}) $
C) $ c^{2}=r^{2}({{\lambda }^{2}}+{{\mu }^{2}}) $
D) $ r^{2}=c^{2}({{\lambda }^{2}}+{{\mu }^{2}}) $
Show Answer
Answer:
Correct Answer: C
Solution:
Polar is $ \lambda x+\mu y+c=0 $ . The condition of tangency $ p=r $ gives the result (c).
BETA
