Question

If the variable line $3x-4y+k=0$ lies between the circles $x^2+y^2-2x-2y+1=0and{x}^2+y^2-16x-2y+61=0$ without intersecting or touching either circle, then the range of $k$ is $(a,b)$, where $a,b,\in I.$ Find the value of $(b-a)$.

Answer 1

The given circles:

$C_1(x-1{)}^2+(y-1{)}^2=1$

$C_2(x-8{)}^2+(y-1{)}^2=4$

$3x-4y+k=0$

$(3\cdot 1-4\cdot 1+k)(3\cdot 8-4\cdot 1+k)<0\Rightarrow (k-1)(k+20)<0\Rightarrow k\in (-20,1)$

For $C_1$

$\frac{|3\cdot 1-4\cdot 1+k|}{5}>1\Rightarrow \frac{|k-1|}{5}>1$

$\Rightarrow k-1>5$ or $k-1<5$

$\Rightarrow k>6$ or $k<-4$

For $C_2$

$\frac{|3\cdot 8-4\cdot 1+k|}{5}>2\Rightarrow \frac{|k+20|}{5}>2$

$\Rightarrow k+20>10$ or $k+20<-10$

$\Rightarrow k>-10$ or $k<-30$

$\Rightarrow k\in (-10,-4)\Rightarrow a=-10,b=-4$

$\Rightarrow b-a=-4+10=6$