© All Rights reserved @ LearnWithDash
Step-by-Step Solution
1. Write down the equations of the circles and the line
• Circle 1: (x - 1)^2 + (y - 1)^2 = 1 , with center C_1(1,\,1) and radius r_1 = 1 .
• Circle 2: (x - 9)^2 + (y - 1)^2 = 4 , with center C_2(9,\,1) and radius r_2 = 2 .
• Line: 3x + 4y = \alpha , or 3x + 4y - \alpha = 0 .
2. Determine the condition for a line not to intersect a circle
For a line Ax + By + C = 0 and a circle with center (h,\,k) and radius r , the line does not intersect the circle if the perpendicular distance of the center from the line is strictly greater than r . The perpendicular distance d of the center from the line is given by:
d = \dfrac{|Ah + Bk + C|}{\sqrt{A^2 + B^2}} > r.
3. Apply the perpendicular distance condition for Circle 1
Circle 1 has center C_1(1,\,1) and radius 1 . The line is 3x + 4y - \alpha = 0 . Thus,
d_1 = \dfrac{|3(1) + 4(1) - \alpha|}{\sqrt{3^2 + 4^2}}
= \dfrac{|7 - \alpha|}{5}.
For no intersection:
\dfrac{|7 - \alpha|}{5} > 1 \quad \Longrightarrow \quad |7 - \alpha| > 5.
This gives two inequalities:
1) 7 - \alpha > 5 \quad \Longrightarrow \quad \alpha < 2,
2) \alpha - 7 > 5 \quad \Longrightarrow \quad \alpha > 12.
Hence, for Circle 1: \alpha < 2 or \alpha > 12.
4. Apply the perpendicular distance condition for Circle 2
Circle 2 has center C_2(9,\,1) and radius 2 . The line is still 3x + 4y - \alpha = 0 . Thus,
d_2 = \dfrac{|3(9) + 4(1) - \alpha|}{\sqrt{3^2 + 4^2}}
= \dfrac{|27 + 4 - \alpha|}{5}
= \dfrac{|31 - \alpha|}{5}.
For no intersection:
\dfrac{|31 - \alpha|}{5} > 2 \quad \Longrightarrow \quad |31 - \alpha| > 10.
This gives two inequalities:
1) 31 - \alpha > 10 \quad \Longrightarrow \quad \alpha < 21,
2) \alpha - 31 > 10 \quad \Longrightarrow \quad \alpha > 41.
Hence, for Circle 2: \alpha < 21 or \alpha > 41.
5. Combine the conditions and interpret “lying between the two circles”
The line must not intersect either circle and must lie in the region strictly between them. From the centers and radii, the tangent lines facing each other (the “inner” tangents) occur at:
Circle 1 on the side toward Circle 2: \alpha = 12 .
Circle 2 on the side toward Circle 1: \alpha = 21 .
Thus, to be “between” the two circles without intersecting either, we need
12 < \alpha < 21.
6. Find the integral values of α and their sum
All integer \alpha in the range 12 < \alpha < 21 are:
\alpha = 13, 14, 15, 16, 17, 18, 19, 20.
Their sum is
13 + 14 + 15 + 16 + 17 + 18 + 19 + 20.
We can group or use the formula for sum of consecutive integers:
S = 13 + 14 + 15 + 16 + 17 + 18 + 19 + 20 = 8 \times \dfrac{(13 + 20)}{2} = 4 \times 33 = 132.
7. Final Answer
The sum of all integral values of \alpha for which the line lies strictly between the two circles without intersecting them is
\boxed{132}.
