If the number of common tangents to the pair | vedclass

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(B) Given the equations of the circles: $x^2+y^2-2x+4y-4=0$ $x^2+y^2+4x-4y+\alpha=0$ For the first circle,the center $C_1 = (1, -2)$ and radius $r_1 =

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$5$

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(B) Given the equations of the circles: $x^2+y^2-2x+4y-4=0$ $x^2+y^2+4x-4y+\alpha=0$ For the first circle,the center $C_1 = (1, -2)$ and radius $r_1 = \sqrt{1^2 + (-2)^2 - (-4)} = \sqrt{1+4+4} = 3$. For the second circle,the center $C_2 = (-2, 2)$ and radius $r_2 = \sqrt{(-2)^2 + 2^2 - \alpha} = \sqrt{8-\alpha}$. For two circles to have $4$ common tangents,they must be separated,which means the distance between their centers $d = C_1C_2$ must be greater than the sum of their radii: $d > r_1 + r_2$ $d = \sqrt{(1 - (-2))^2 + (-2 - 2)^2} = \sqrt{3^2 + (-4)^2} = \sqrt{9+16} = 5$. So,$5 > 3 + \sqrt{8-\alpha}$ $2 > \sqrt{8-\alpha}$ Squaring both sides (since both sides are positive): $4 > 8 - \alpha$ $\alpha > 4$. The least integral value of $\alpha$ greater than $4$ is $5$.

If the number of common tangents to the pair of circles $x^2+y^2-2x+4y-4=0$ and $x^2+y^2+4x-4y+\alpha=0$ is $4$,then the least integral value of $\alpha$ is

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