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 $\lambda$ will lie in the interval:

• A. $(23,31)$
• B. $(2,17)$
• C. $[13,23]$
• D. $[12,21]$

Answer 1

The correct option is D $[12,21]$

Let the first circle be $C_1:x^2+y^2-2x-2y+1=0$
Centre $\equiv O(1,1)$
Radius $\equiv r_1=1$
and the second circle be $C_2:x^2+y^2-18x-2y+78=0$
Centre $\equiv O^{\prime }(9,1)$
Radius $\equiv r_2=2$
$O{O}^{\prime }=8>r_1+r_2$



Now, both circles lies on opposite side of line $3x+4y-\lambda =0$, we get
$(3+4-\lambda )(27+4-\lambda )<0\Rightarrow (7-\lambda )(31-\lambda )<0\Rightarrow (\lambda -7)(\lambda -31)<0\Rightarrow \lambda \in (7,31)\cdots \left(i\right)$

Now,
$OP\ge r_1\Rightarrow \frac{|3+4-\lambda |}{5}\ge 1\Rightarrow |7-\lambda |\ge 5\Rightarrow \lambda \ge 12\text{or}\lambda \le 2\therefore \lambda \in (-\infty ,2]\cup [12,\infty )\cdots \left(ii\right)$

And
$O^{\prime }P\ge r_2$
$\Rightarrow \frac{|27+4-\lambda |}{5}\ge 2$
$\Rightarrow \lambda \ge 41\text{or}\lambda \le 21\therefore \lambda \in (-\infty ,21]\cup [41,\infty )\cdots \left(iii\right)$

From $\left(i\right),\left(ii\right)$ and $\left(iii\right)$
$\lambda \in [12,21]$