A) \[2x+y-1=0\]
B) \[2x+y+1=0\]
C) \[x+2y-1=0\]
D) \[x+2y+1=0\]
Correct Answer: C
Solution :
Any line parallel to \[x+2y=3\] is \[x+2y+\lambda =0\] and for this to be a normal to the given circle, must pass through its centre (1, 0), i.e. \[\lambda =-1\]. So, normal is \[x+2y-1=0\].You need to login to perform this action.
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