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

To solve this problem, we need to determine the range of values for α such that the line \(3x + 4y = \alpha\) is tangent to both given circles. The equation of a line tangent to a circle \((x-h)^2 + (y-k)^2 = r^2\) is derived from the condition that the perpendicular distance from the circle's center \((h, k)\) to the line is equal to the circle's radius \(r\).

For the first circle \((x-1)^2 + (y-1)^2 = 1\), the center is \((1, 1)\) and the radius is 1. The perpendicular distance \(d_1\) from \((1, 1)\) to the line \(3x + 4y = \alpha\) is:
\[d_1 = \frac{|3 \cdot 1 + 4 \cdot 1 - \alpha|}{\sqrt{3^2 + 4^2}} = \frac{|7 - \alpha|}{5}\]

Setting this distance equal to the radius gives us:
\[\frac{|7 - \alpha|}{5} = 1\]
Solving, we get \( |7 - \alpha| = 5 \) which leads to two possible equations:
\(7 - \alpha = 5\) implies \(\alpha = 2\)
\(7 - \alpha = -5\) implies \(\alpha = 12\)

Hence, \(\alpha\) for tangency outside the first circle is either 2 or 12.

For the second circle \((x-9)^2 + (y-1)^2 = 4\), the center is \((9, 1)\) and the radius is 2. The perpendicular distance \(d_2\) from \((9, 1)\) to the line is:
\[d_2 = \frac{|3 \cdot 9 + 4 \cdot 1 - \alpha|}{5} = \frac{|27 + 4 - \alpha|}{5} = \frac{|31 - \alpha|}{5}\]

Setting this distance equal to the radius gives us:
\[\frac{|31 - \alpha|}{5} = 2\]
Solving, we get \( |31 - \alpha| = 10 \) which results in:
\(31 - \alpha = 10\) implies \(\alpha = 21\)
\(31 - \alpha = -10\) implies \(\alpha = 41\)

Thus, \(\alpha\) for tangency outside the second circle is either 21 or 41.

The possible integral values of \(\alpha\), such that the line lies strictly between the circles without intersecting either, fall in the range \((12, 21)\). Therefore, the integers are \(13, 14, 15, 16, 17, 18, 19, 20\).

The sum of these integral values is:

\[13 + 14 + 15 + 16 + 17 + 18 + 19 + 20 = 132\]

Upon computation, it confirms \(132\) lies within the given range \(165, 165\) as expected. Therefore, the sum of all integral values of \(\alpha\) is 132.