Question

The range of parameter $a$ for which the variable line $y=2x+a$ lies between the circles $x^2+y^2-2x-2y+1=0$ and $x^2+y^2-16x-2y+61=0$ without intersecting or touching either circle is

• A. $(-15,-1)$
• B. $(-\infty ,-2\sqrt{5}-15)$
• C. $(2\sqrt{5}-15,-\sqrt{5}-1)$
• D. $(\sqrt{5}-1,\infty )$

Answer 1

The correct option is C $(2\sqrt{5}-15,-\sqrt{5}-1)$

Given circles are $S_1:x^2+y^2-2x-2y+1=0$ and $S_2:x^2+y^2-16x-2y+61=0$
The centre and radius are
$C_1=(1,1),r_1=1C_2=(8,1),r_2=2$
Given line is $y=2x+a\Rightarrow 2x-y+a=0$


The line lies between these circles if centre of the circles lies on opposite sides of the line, we get
$\Rightarrow (2-1+a)(16-1+a)<0\Rightarrow a\in (-15,-1)\cdots \left(1\right)$

Line will not touch or intersect the circles, if $d>r$, where $d$ is perpendicular distance from the center of circle to the line

Now for circle $S_1$, we get
$\frac{|a+1|}{\sqrt{5}}>1\Rightarrow a>\sqrt{5}-1\text{or}a<-\sqrt{5}-1\Rightarrow a\in (-\infty ,-\sqrt{5}-1)\cup (\sqrt{5}-1,\infty )\dots \left(2\right)$

And for circle $S_2$, we get
$\frac{|a+15|}{\sqrt{5}}>2$
$\Rightarrow a>2\sqrt{5}-15$ or $a<-2\sqrt{5}-15$
$\Rightarrow a\in (-\infty ,-2\sqrt{5}-15)\cup (2\sqrt{5}-15,\infty )\dots \left(3\right)$

Hence, final solution set of $a$ is
$\left(1\right)\cap \left(2\right)\cap \left(3\right)$
$\Rightarrow a\in (2\sqrt{5}-15,-\sqrt{5}-1)$

Alternate Solution:
Using tangency condition for $C_1$ with $y=2x+a$
$\frac{|a+1|}{\sqrt{5}}=1\Rightarrow a=\pm \sqrt{5}-1$
from the image $y$ intercept of the line is negative
$\therefore a=-\sqrt{5}-1\cdots \left(1\right)$
Using tangency condition for $C_1$ with $y=2x+a$
$\frac{|a+15|}{\sqrt{5}}=2\Rightarrow a=\pm 2\sqrt{5}-15$
From the image it is clear that $y$ intercept of the line should be higher among $\pm 2\sqrt{5}-15$
So, $a=2\sqrt{5}-15\cdots \left(2\right)$
Now from $\left(1\right)$ and $\left(2\right)$, for the line to lie in between the circles
$a\in (2\sqrt{5}-15,-\sqrt{5}-1)$