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```        The value of m for which the line y = mx lies wholly outside the circle x2 + y2 - 2x - 4y + 1 = 0 is / are
(a) (-4 /3 , 0 )         (b) (-4/3 , 0]          (c) (0, 4/3)        (d) none```
8 years ago

147 Points
```										Dear Sanchit
x2 + y2 - 2x - 4y + 1 = 0
compair with standard question
perpendicular distance from center to line y=mx must be greater than radius of circle
mod[(m*1-1*2)/√(1+m2)]  >2
so (m-2)2 >4(1+m2)
m2 +4 -4m >4 +4m2
3m2+4m <0
m(3m+4)<0
option a is correct
Please feel free to post as many doubts on our discussion forum as you can. If you find any question Difficult to understand - post it here and we will get you the answer and detailed solution very quickly. We are all IITians and here to help you in your IIT JEE preparation. All the best. Regards,Askiitians ExpertsBadiuddin

```
8 years ago
23 Points
```										Dear Sanchit,
Given the circle equation as x2 + y2 - 2x - 4y + 1 = 0
Its center is (1,2) and radius is 2
Now the given line to be wholely outside the circle its perpendicular distance from the centre of the circle should be greater than the radius of the circle
thus perpendicular distance of line y=mx from (1,2) is
d = modulus of{(m-1)/√(m2+1)} And the condition is d > the radius of circle
---->modulus of{(m-1)/√(m2+1)} >2
By squaring on both sides and simplifying we get
---> m2 + (4/3)m  < 0
Thus
-(4/3) < m <0
is the range of m and hence
Please feel free to post as many doubts on our discussion forum as you can. If you find any question Difficult to understand - post it here and we will get you the answer and detailed solution very quickly. We are all IITians and here to help you in your IIT JEE preparation.
All the best  !!!

Regards,
```
8 years ago
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