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

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, $\frac{\left|3 \times 1 - 4 \times 1 + k\right|}{\sqrt{3^{2} + 4^{2}}}>1\&\frac{\left|3 \times 8 - 4 \times 1 + k\right|}{\sqrt{3^{2} + 4^{2}}}>2$
$\Rightarrow \frac{\left|k - 1\right|}{5}>1\&\frac{\left|k + 20\right|}{5}>2$
$\Rightarrow \left|k - 1\right|>5\&\left|k + 20\right|>10$
$\Rightarrow \left(k - 1 > 5 \text{or} k - 1 < - 5\right)\&\left(k + 20 > 10 \text{or} k + 20 < - 10\right)$
$\Rightarrow \left(k > 6 \text{or} k < - 4\right)\&\left(k > - 10 \text{or} k < - 30\right)$
$\Rightarrow k\in \left(- \infty , - 30\right)\cup\left(- 10 , - 4\right)\cup\left(6 , \infty \right)..............\left(i i\right)$
So, the range of $k$ is the intersection of both the equations.
$\therefore k\in \left(- 10 , - 4\right)$
So, $\left(a , b\right)=\left(- 10 , - 4\right)\Rightarrow a=-10\&b=-4$
$\therefore b-a=\left(- 4\right)-\left(- 10\right)=6$ .