Circle And System Of Circles Question 306
Question: The equations of the tangents drawn from the point (0, 1) to the circle $ x^{2}+y^{2}-2x+4y=0 $ are
[Roorkee 1979]
Options:
A) $ 2x-y+1=0,x+2y-2=0 $
B) $ 2x-y+1=0,x+2y+2=0 $
C) $ 2x-y-1=0,x+2y-2=0 $
D) $ 2x-y-1=0,x+2y+2=0 $
Show Answer
Answer:
Correct Answer: A
Solution:
Required equations are given by $ SS_1=T^{2} $
$ \Rightarrow (x^{2}+y^{2}-2x+4y)(1+4)={{{y-1(x)+2(y+1)}}^{2}} $
$ \Rightarrow 2x^{2}-2y^{2}-3x+4y+3xy-2=0 $
$ \Rightarrow (2x-y+1)(x+2y-2)=0 $ .
BETA
