 # 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 Badiuddin askIITians.ismu Expert
148 Points
13 years ago

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

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13 years ago

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

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