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Sanchit Gupta Grade: 12
        

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

7 years ago

Answers : (2)

Badiuddin askIITians.ismu Expert
147 Points
										

Dear Sanchit


x2 + y2 - 2x - 4y + 1 = 0


compair with standard question


center(1,2)  radius  2


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


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Badiuddin





7 years ago
Askiitians Expert Bharath-IITD
23 Points
										

Dear Sanchit,


Given the circle equation as x2 + y- 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


answer is (a)


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 Experts


Adapa Bharath

7 years ago
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