If the circles x^2+y^2+2 lambda x+2=0 and x^2 | vedclass

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(B) The centres of the two circles are $C_1(-\lambda, 0)$ and $C_2(0, -2)$ and their radii are $r_1 = \sqrt{\lambda^2-2}$ and $r_2 = \sqrt{2}$.The two

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$\pm 2$

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(B) The centres of the two circles are $C_1(-\lambda, 0)$ and $C_2(0, -2)$ and their radii are $r_1 = \sqrt{\lambda^2-2}$ and $r_2 = \sqrt{2}$.The two circles touch each other if the distance between their centres $C_1C_2$ is equal to the sum or difference of their radii: $C_1C_2 = |r_1 \pm r_2|$.Calculating the distance $C_1C_2 = \sqrt{(-\lambda - 0)^2 + (0 - (-2))^2} = \sqrt{\lambda^2 + 4}$.Setting $C_1C_2 = r_1 + r_2$:$\sqrt{\lambda^2 + 4} = \sqrt{\lambda^2 - 2} + \sqrt{2}$Squaring both sides:$\lambda^2 + 4 = (\lambda^2 - 2) + 2 + 2\sqrt{2(\lambda^2 - 2)}$$\lambda^2 + 4 = \lambda^2 + 2\sqrt{2(\lambda^2 - 2)}$$4 = 2\sqrt{2(\lambda^2 - 2)}$$2 = \sqrt{2(\lambda^2 - 2)}$Squaring again:$4 = 2(\lambda^2 - 2)$$2 = \lambda^2 - 2$$\lambda^2 = 4 \Rightarrow \lambda = \pm 2$.

If the circles $x^2+y^2+2 \lambda x+2=0$ and $x^2+y^2+4y+2=0$ touch each other,then $\lambda=$

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