1. Given circles and line

We have the circles:

$C_1:(x-1)^2+(y-1)^2=1$ Center $O_1=(1,1)$, radius $r_1=1$.

$C_2:(x-9)^2+(y-1)^2=4$ Center $O_2=(9,1)$, radius $r_2=2$.

And the variable line:

$3x+4y=\alpha$

We want the line to lie between the two circles and not cut a chord on either circle.

That means:

• it must not intersect either circle,
• and it should be located in the region separating them.

So the line must be an external common tangent-like separator between the circles.

2. Condition for not intercepting a chord

A line does not cut a chord of a circle if its distance from the center is at least the radius.

Distance of line $3x+4y=\alpha$ from point $(x_0,y_0)$ is

$\frac{|3x_0+4y_0-\alpha|}{\sqrt{3^2+4^2}}=\frac{|3x_0+4y_0-\alpha|}{5}$

For $C_1$

Center $(1,1)$:

$\frac{|3(1)+4(1)-\alpha|}{5}=\frac{|7-\alpha|}{5}$

For no chord:

$\frac{|7-\alpha|}{5}\ge 1$

$|7-\alpha|\ge 5$

So,

$\alpha\le 2 \quad \text{or} \quad \alpha\ge 12$

For $C_2$

Center $(9,1)$:

$\frac{|3(9)+4(1)-\alpha|}{5}=\frac{|31-\alpha|}{5}$

For no chord:

$\frac{|31-\alpha|}{5}\ge 2$

$|31-\alpha|\ge 10$

So,

$\alpha\le 21 \quad \text{or} \quad \alpha\ge 41$

3. Condition that the line lies between the circles

For a line to lie between the two circles, the two centers must lie on opposite sides of the line.

Evaluate $3x+4y$ at the centers:

• At $(1,1)$: $3+4=7$
• At $(9,1)$: $27+4=31$

Thus the line $3x+4y=\alpha$ lies between them when

$7<\alpha<31$

(If $\alpha$ were outside this interval, both centers would be on the same side.)

4. Combine all conditions

We need:

$7<\alpha<31$

and also

$\alpha\le 2 \text{ or } \alpha\ge 12$

and

$\alpha\le 21 \text{ or } \alpha\ge 41$

Within $7<\alpha<31$:

• from first circle, only $\alpha\ge 12$ survives,
• from second circle, only $\alpha\le 21$ survives.

Hence,

$12\le \alpha\le 21$

These are exactly the values for which the line stays between the circles and does not cut a chord on either circle.

5. Integral values of $\alpha$

Integral values are:

$12,13,14,15,16,17,18,19,20,21$

Their sum is

$\frac{(12+21)\cdot 10}{2}=\frac{33\cdot 10}{2}=165$

6. Final answer

$\boxed{165}$

The derived answer matches the stored correct answer.