If the variable line 3 x + 4 y = α lies between the two circles ( x - 1 ) 2 + ( y - 1 ) 2 = 1 and ( x -…

If the variable line 3x+4y=α 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 α is

Solution

Given line $3x + 4y = \alpha$ Given circles $(x-1)^2 + (y-1)^2 = 1$ and $(x-9)^2 + (y-1)^2 = 4$ According to the given information centres of both the circles must lie on opposite side of line. $L_{11} \cdot L_{22} < 0$ $\Rightarrow (3 + 4 - \alpha)(27 + 4 - \alpha) < 0$ $\alpha \in (7, 31)$ Line is neither touching nor intersecting that means perpendicular distance from centre to line must more than or equal to radius. $\Rightarrow \frac{|3 + 4 - \alpha|}{5} \geq 1$ and $\frac{|27 + 4 - \alpha|}{5} \geq 2$ $\Rightarrow \alpha \in (-\infty, 2] \cup [12, \infty)$ and $\alpha \in (-\infty, 21] \cup [41, \infty)$ $\Rightarrow \alpha \in [12, 21]$ Sum of integer values of $\alpha = 12 + 13 + \dots + 21 = 165$

Asked in: JEE Main 2021 (31 Aug Shift 1)

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