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

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

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

answer is (a)

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Adapa Bharath