If a variable line,3x + 4y - lambda = 0,is su | vedclass

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(C) The centers of the circles are $C_1(1, 1)$ and $C_2(9, 1)$.Their radii are $r_1 = \sqrt{1^2 + 1^2 - 1} = 1$ and $r_2 = \sqrt{9^2 + 1^2 - 78} = \sq

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$[12, 21]$

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(C) The centers of the circles are $C_1(1, 1)$ and $C_2(9, 1)$.Their radii are $r_1 = \sqrt{1^2 + 1^2 - 1} = 1$ and $r_2 = \sqrt{9^2 + 1^2 - 78} = \sqrt{81 + 1 - 78} = 2$.For the circles to lie on opposite sides of the line $L: 3x + 4y - \lambda = 0$,the values of the expression $f(x, y) = 3x + 4y - \lambda$ at the centers must have opposite signs,and the line must not intersect the circles.$f(C_1) = 3(1) + 4(1) - \lambda = 7 - \lambda$.$f(C_2) = 3(9) + 4(1) - \lambda = 31 - \lambda$.Condition for opposite sides: $(7 - \lambda)(31 - \lambda) Also,the distance from the center to the line must be at least the radius:$d_1 = \frac{|3(1) + 4(1) - \lambda|}{\sqrt{3^2 + 4^2}} = \frac{|7 - \lambda|}{5} \ge 1 \implies |7 - \lambda| \ge 5 \implies \lambda \le 2$ or $\lambda \ge 12$.$d_2 = \frac{|3(9) + 4(1) - \lambda|}{\sqrt{3^2 + 4^2}} = \frac{|31 - \lambda|}{5} \ge 2 \implies |31 - \lambda| \ge 10 \implies \lambda \le 21$ or $\lambda \ge 41$.Taking the intersection of $\lambda \in (7, 31)$,$\lambda \in (-\infty, 2] \cup [12, \infty)$,and $\lambda \in (-\infty, 21] \cup [41, \infty)$,we get $\lambda \in [12, 21]$.

If a variable line,$3x + 4y - \lambda = 0$,is such that the two circles $x^2 + y^2 - 2x - 2y + 1 = 0$ and $x^2 + y^2 - 18x - 2y + 78 = 0$ are on its opposite sides,then the set of all values of $\lambda$ is the interval

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