Circle And System Of Circles Question 268
Question: The equations of the tangents drawn from the origin to the circle $ x^{2}+y^{2}-2rx-2hy+h^{2}=0 $ are
[Roorkee 1989; IIT 1988; RPET 1996]
Options:
A) $ x=0,y=0 $
B) $ (h^{2}-r^{2})x-2rhy=0,x=0 $
C) $ y=0,x=4 $
D) $ (h^{2}-r^{2})x+2rhy=0,x=0 $
Show Answer
Answer:
Correct Answer: B
Solution:
The equation of tangents is $ SS_1=T^{2} $
$ \Rightarrow h^{2}(x^{2}+y^{2}-2rx-2hy+h^{2})={{(rx+hy-h^{2})}^{2}} $
$ \Rightarrow (h^{2}-r^{2})x^{2}-2rhxy=0\Rightarrow x{(h^{2}-r^{2})x-2rhy}=0 $
$ \Rightarrow x=0,\ (h^{2}-r^{2})x-2rhy=0 $ .
BETA
