Mathematics - Circle Question with Solution | TestHub

Given line $3x + 4y = \alpha$

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.

$L_{11} . L_{22} < 0$

$\Rightarrow \left( 3 +4 -\alpha \right) \left(27 + 4 - \alpha \right) <0$

$\alpha \in \left(7, 31 \right)$.

Line is neither touching nor intersecting that means perpendicular distance from  centre to line must more than or equal to radius.

$\Rightarrow \frac{\left|3+4-\alpha \right|}{5} \ge 1 \& \frac{\left|27+4-\alpha \right|}{5} \ge 2$

$\Rightarrow \alpha \in (-\infty , 2] \cup [12,\infty ) \& \alpha \in (-\infty , 21] \cup [41,\infty )$

$\Rightarrow \alpha \in \left[12, 21\right]$

Sum of integer values of $\alpha = 12 + 13+ --21$$=165.$