Circle And System Of Circles Question 263
Question: If the straight line $ y=mx $ is outside the circle $ x^{2}+y^{2}-20y+90=0 $ , then
[Roorkee 1999]
Options:
A) $ m>3 $
B) $ m<3 $
C) $ |m|>3 $
D) $ |m|<3 $
Show Answer
Answer:
Correct Answer: D
Solution:
If the straight line y = mx is outside the given circle then distance from centre of circle > radius of circle $ \frac{10}{\sqrt{1+m^{2}}}>\sqrt{10} $
therefore (1 + m2) < 10
therefore m2 < 9
therefore |m|< 3.
BETA
