If the variable line 3x + 4y = alpha lies bet | vedclass

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(C) The line $3x + 4y - \alpha = 0$ lies between the two circles if the centers $(1, 1)$ and $(9, 1)$ lie on opposite sides of the line.Substituting t

Vedclass · Accepted Answer

$165$

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(C) The line $3x + 4y - \alpha = 0$ lies between the two circles if the centers $(1, 1)$ and $(9, 1)$ lie on opposite sides of the line.Substituting the centers into the line equation: $f(1, 1) = 3(1) + 4(1) - \alpha = 7 - \alpha$ and $f(9, 1) = 3(9) + 4(1) - \alpha = 31 - \alpha$.For the centers to be on opposite sides,$(7 - \alpha)(31 - \alpha) Also,the line should not intersect either circle,meaning the distance from the center to the line must be greater than or equal to the radius.For circle $1$: $d_1 = \frac{|3(1) + 4(1) - \alpha|}{\sqrt{3^2 + 4^2}} = \frac{|7 - \alpha|}{5} \geq 1$ $\Rightarrow |7 - \alpha| \geq 5$ $\Rightarrow \alpha \leq 2$ or $\alpha \geq 12$.For circle $2$: $d_2 = \frac{|3(9) + 4(1) - \alpha|}{\sqrt{3^2 + 4^2}} = \frac{|31 - \alpha|}{5} \geq 2$ $\Rightarrow |31 - \alpha| \geq 10$ $\Rightarrow \alpha \leq 21$ or $\alpha \geq 41$.Combining the conditions: $\alpha \in (7, 31) \cap ((\infty, 2] \cup [12, \infty)) \cap ((\infty, 21] \cup [41, \infty)) = [12, 21]$.The integral values of $\alpha$ are $12, 13, 14, 15, 16, 17, 18, 19, 20, 21$.The sum is $\frac{10}{2}(12 + 21) = 5 	imes 33 = 165$.

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 .... .

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