Circle And System Of Circles Question 348
Question: If the straight line $ y=mx+c $ touches the circle $ x^{2}+y^{2}-4y=0 $ , then the value of c will be
[RPET 1988]
Options:
A) $ 1+\sqrt{1+m^{2}} $
B) $ 1-\sqrt{m^{2}+1} $
C) $ 2(1+\sqrt{1+m^{2}}) $
D) $ 2+\sqrt{1+m^{2}} $
Show Answer
Answer:
Correct Answer: C
Solution:
Apply for tangency of line, centre being (0, 2) and radius = 2 $ | \frac{-2+c}{\sqrt{1+m^{2}}} |=2\Rightarrow c^{2}-4c+4=4+4m^{2} $
$ \Rightarrow c=\frac{4\pm \sqrt{16+16m^{2}}}{2} $ or $ c=2\pm 2\sqrt{1+m^{2}} $ Most correct answer is $ c=2(1+\sqrt{1+m^{2}}) $ .
BETA
