JEE Main 12 Jan 2019 Morning Question 16
Question: If a variable line, $ 3x+4y-\lambda =0 $ is such that the two circles $ x^{2}+y^{2}-2x-2y+1=0 $ and $ x^{2}+y^{2}-18x $ $ -2y+78=0 $ are on its opposite sides, then the set of all values of $ \lambda $ is the interval [JEE Main Online Paper Held On 12-Jan-2019 Morning]
Options:
A) (23, 31)
B) (2, 17)
C) [13, 23]
D) [12, 21]
Show Answer
Answer:
Correct Answer: D
Solution:
Let $ S _1:x^{2}+y^{2}-2x-2y+1=0 $ $ C _1(1,1),r _1=1 $ and $ S _2:x^{2}+y^{2}-18x-2y+78=0 $ $ C _2(9,1),r _2=2 $
Centre of circles lie on opposite side of line $ 3x+4y-\lambda =0 $
$ \therefore $ $ (3+4-\lambda )(27+4-\lambda )<0 $
$ \Rightarrow $ $ (\lambda -7)(\lambda -31)<0 $
$ \Rightarrow $ $ \lambda \in (7,31) $ Line lies outside the circles $ S _1 $ and $ S _2. $
$ \therefore $ $ | \frac{3+4-\lambda }{5} |\ge 1\Rightarrow \lambda \in (-\infty ,2]\cup [12,\infty ) $ and $ | \frac{27+4-\lambda }{5} |\ge 2\Rightarrow \lambda \in (-\infty ,21]\cup [41,\infty ) $ So, $ \lambda \in [12,21] $
