Question

If the variable line $3x+4y=\alpha$ lies between the two circles $(x-1{)}^2+(y-1{)}^2=1$ and $(x-9{)}^2+(y-1{)}^2=4,$ without intercepting a chord on either circle, then the sum of all the integral values of $\alpha$ is

Answer 1

$C_1=(1,1)\text{and}{r}_1=1$
$C_2=(9,1)\text{and}{r}_2=2$
$L\equiv 3x+4y-\alpha =0$
Distance of line from $C_i$ should be greater than $r_i(i=1,2)$
$\Rightarrow \left|\frac{7-\alpha }{5}\right|>1$
$\Rightarrow |\alpha -7|>5$
$\Rightarrow \alpha \in (-\infty ,2)\cup (12,\infty )\cdots \left(i\right)$
Also,
$\left|\frac{27+4-\alpha }{5}\right|>2\Rightarrow |\alpha -31|>10\Rightarrow \alpha \in (-\infty ,21)\cup (41,\infty )\cdots \left(ii\right)$
Further $C_1$ and $C_2$ should lie on opposite sides w.r.t given lines
$\Rightarrow (3+4-\alpha )\cdot (27+4-\alpha )<0\Rightarrow (\alpha -7)(\alpha -31)<0\Rightarrow \alpha \in (7,31)\cdots \left(iii\right)$
From $\left(i\right),\left(ii\right)$ and $\left(iii\right)$
$\alpha \in (12,21)$
Sum of all the integral values of $\alpha$
$=12+13+14+\cdots +21=\frac{10}{2}[12+21]=165$