Thank you for registering.

One of our academic counsellors will contact you within 1 working day.

Please check your email for login details.
MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping

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

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

Grade:12

2 Answers

Badiuddin askIITians.ismu Expert
147 Points
11 years ago

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

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
Badiuddin



Askiitians Expert Bharath-IITD
23 Points
11 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

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

Think You Can Provide A Better Answer ?

Provide a better Answer & Earn Cool Goodies See our forum point policy

ASK QUESTION

Get your questions answered by the expert for free