Given line 3x+4y=α
Given circles (x-1{)}^{2}+(y-1{)}^{2}=1&(x-9{)}^{2}+(y-1{)}^{2}=4
According to the given information centres of both the circles must lie on opposite side of line.
L11.L22<0
⇒(3+4−α)(27+4−α)<0
α∈(7,31).
Line is neither touching nor intersecting that means perpendicular distance from centre to line must more than or equal to radius.
\Rightarrow \frac{|3+4-\alpha |}{5}\geq 1&\frac{|27+4-\alpha |}{5}\geq 2
\Rightarrow \alpha \in (-\infty ,2]\cup [12,\infty )&\alpha \in (-\infty ,21]\cup [41,\infty )
⇒α∈[12,21]
Sum of integer values of α=12+13+−−21=165.