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 _____________
 

Hint:

The "opposite sides" condition is mathematically expressed by the product of the line's values at the two points being negative. This significantly narrows down the possible range for \(\alpha\).

Answer 1

Correct Answer: 165

Step 1: Understanding the Concept:
For a line to lie between two circles without intersecting them, its distance from the centers must be greater than or equal to the respective radii, and the centers must lie on opposite sides of the line.
Step 2: Detailed Explanation:
Circle 1: Center \(C_1(1,1)\), Radius \(r_1 = 1\).
Circle 2: Center \(C_2(9,1)\), Radius \(r_2 = 2\).
Distance of line \(3x+4y-\alpha = 0\) from centers:
1. \(d_1 \geq 1 \implies \frac{|3(1)+4(1)-\alpha|}{5} \geq 1 \implies |7-\alpha| \geq 5 \implies \alpha \leq 2\) or \(\alpha \geq 12\).
2. \(d_2 \geq 2 \implies \frac{|3(9)+4(1)-\alpha|}{5} \geq 2 \implies |31-\alpha| \geq 10 \implies \alpha \leq 21\) or \(\alpha \geq 41\).
3. Since the line is between the circles, the centers must be on opposite sides:
\[ (3(1)+4(1)-\alpha)(3(9)+4(1)-\alpha)<0 \implies (7-\alpha)(31-\alpha)<0 \implies 7<\alpha<31 \]
Combining all conditions:
- From \(7<\alpha<31\) and (\(\alpha \leq 2\) or \(\alpha \geq 12\)): \(\alpha \in [12, 31)\).
- From \(\alpha \in [12, 31)\) and (\(\alpha \leq 21\) or \(\alpha \geq 41\)): \(\alpha \in [12, 21]\).
Integral values of \(\alpha = \{12, 13, 14, 15, 16, 17, 18, 19, 20, 21\}\).
Sum = \(\frac{10}{2}(12 + 21) = 5 \times 33 = 165\).
Step 3: Final Answer:
The sum of integral values is 165.