Circle And System Of Circles Question 372
Question: The line $ y=mx+c $ intersects the circle $ x^{2}+y^{2}=r^{2} $ at two real distinct points, if
Options:
A) $ -r\sqrt{1+m^{2}}<c\le 0 $
B) $ 0\le c<r\sqrt{1+m^{2}} $
C) (a) and (b) both
D) $ -c\sqrt{1-m^{2}}<r $
Show Answer
Answer:
Correct Answer: C
Solution:
Substituting equation of line $ y=mx+c $ in circle $ x^{2}+y^{2}=r^{2} $
$ x^{2}+{{(mx+c)}^{2}}=r^{2}\Rightarrow {{(1+m)}^{2}}x^{2}+2mxc+c^{2}-r^{2}=0 $ If discriminant is greater than zero; two real values of x will be obtained so, $ B^{2}>4AC $ . $ 4m^{2}c^{2}-4(c^{2}-r^{2})(1+m^{2})>0 $
$ r^{2}(1+m^{2})>c^{2} $
$ 0\le c<r\sqrt{1+m^{2}} $ and $ -r\sqrt{(1+m^{2})}<c\le 0 $ .
BETA
