Pair Of Straight Lines Question 128
Question: The lines joining the origin to the points of intersection of the line $ y=mx+c $ and the circle $ x^{2}+y^{2}=a^{2} $ will be mutually perpendicular, if
[Roorkee 1977]
Options:
A) $ a^{2}(m^{2}+1)=c^{2} $
B) $ a^{2}(m^{2}-1)=c^{2} $
C) $ a^{2}(m^{2}+1)=c^{2} $
D) $ a^{2}(m^{2}-1)=2c^{2} $
Show Answer
Answer:
Correct Answer: C
Solution:
Making the equation of circle homogeneous with the help of line $ y=mx+c, $ we get $ x^{2}+y^{2}-a^{2}{{( \frac{y-mx}{c} )}^{2}}=0 $ .
$ \Rightarrow c^{2}x^{2}+c^{2}y^{2}-a^{2}y^{2}-a^{2}m^{2}x^{2}+2a^{2}mxy=0 $
$ \Rightarrow (c^{2}-a^{2}m^{2})x^{2}+(c^{2}-a^{2})y^{2}-2a^{2}mxy=0 $ …..(i)
Hence lines represented by (i) are perpendicular, if $ c^{2}-a^{2}m^{2}+c^{2}-a^{2}=0\Rightarrow 2c^{2}=a^{2}(1+m^{2}) $ .
BETA
