Question

If the variable line $3x-4y+k=0$ 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, then range of $k$ is $\left(a,b\right)$ where $a,b\in I$ . Then the value of $\left(b-a\right)$ , is

Answer 1

Answer: 6

Given $C_1:x^2+y^2-2x-2y+1=0\&C_2:x^2+y^2-16x-2y+61=0$
$\Rightarrow C_1:{\left(x-1\right)}^2+{\left(y-1\right)}^2=1^2\&C_2:{\left(x-8\right)}^2+{\left(y-1\right)}^2=2^2$
So, centre of $C_1\&C_2$ are $\left(1,1\right)\&\left(8,1\right)$ respectively and their radii are $1\&2$ respectively.
According to the given condition in question, the centres must lie opposite to line $3x-4y+k=0$ .
So, $\left(3\times 1-4\times 1+k\right)\left(3\times 8-4\times 1+k\right)<0$
$\Rightarrow \left(k-1\right)\left(k+20\right)<0$
$\Rightarrow k\in \left(-20,1\right)...........\left(i\right)$
Also, distance of given line from the centres must be greater than their radii.
So, $\frac{|3\times 1-4\times 1+k|}{\sqrt{3^2+4^2}}>1\&\frac{|3\times 8-4\times 1+k|}{\sqrt{3^2+4^2}}>2$
$\Rightarrow \frac{|k-1|}{5}>1\&\frac{|k+20|}{5}>2$
$\Rightarrow \left|k-1\right|>5\&\left|k+20\right|>10$
$\Rightarrow \left(k-1>5\text{or}k-1<-5\right)\&\left(k+20>10\text{or}k+20<-10\right)$
$\Rightarrow \left(k>6\text{or}k<-4\right)\&\left(k>-10\text{or}k<-30\right)$
$\Rightarrow k\in \left(-\infty ,-30\right)\cup \left(-10,-4\right)\cup \left(6,\infty \right)..............\left(ii\right)$
So, the range of $k$ is the intersection of both the equations.
$\therefore k\in \left(-10,-4\right)$
So, $\left(a,b\right)=\left(-10,-4\right)\Rightarrow a=-10\&b=-4$
$\therefore b-a=\left(-4\right)-\left(-10\right)=6$ .