 # If a variable line, Question:

If a variable line, $3 x+4 y-\lambda=0$ is such that the two circles $x^{2}+y^{2}-2 x-2 y+1=0$ and $x^{2}+y^{2}-18 x-2 y+78=0$ are on its opposite sides, then the set of all values of $\lambda$ is the interval :

1. (1) $(2,17)$

2. (2) $[13,23]$

3. (3) $[12,21]$

4. (4) $(23,31)$

Correct Option: , 3

Solution:

Condition 1: The centre of the two circles are $(1,1)$ and $(9,1)$. The circles are on opposite sides of the line $3 x+4 y-\lambda=0$.

Put $x=1, y=1$ in the equation of line,

$3(1)+4(1)-\lambda=0$

$7-\lambda=0$

Now, put $x=9, y=1$ in the equation of line,

$3(9)+4(1)-\lambda=0$

Then, $(7-\lambda)(27+4-\lambda)<0$

$\Rightarrow \quad(\lambda-7)(\lambda-31)<0$

$\lambda \in(7,31) \ldots(1)$

Condition 2: Perpendicular distance from centre on line $\geq$ radius of circle.

For $x^{2}+y^{2}-2 x-2 y=1$

$\Rightarrow \quad \frac{|3+4-\lambda|}{5} \geq 1$

$\Rightarrow \quad|\lambda-7| \geq 5$

$\Rightarrow \quad \lambda \geq 12$ or $\quad \lambda \Rightarrow 2 \ldots(2)$

For $x^{2}+y^{2}-18 x-2 y+78=0$

$\frac{|27+4-\lambda|}{5} \geq 2$

$\Rightarrow \lambda \geq 41$ or $\lambda \leq 21 \ldots(3)$

Intersection of $(1),(2)$ and $(3)$ gives $\lambda \in[12,21]$.