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If the circles $x^2+y^2+2 \lambda x+2=0$ and $x^2+y^2+4 y+2=0$ touch each other, then $\lambda=$
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The correct answer is:
$\pm 2$
The centres of the two circles are $C_1(-\lambda, 0)$ and $C_2(0,-2)$ and their radii are $\sqrt{\lambda^2-2}$ and $\sqrt{2}$. So, the two circles will touch each other, if
$$
\begin{array}{ll}
C_1 C_2=\text { sum of radii } \\
\Rightarrow & \sqrt{\lambda^2+4}=\sqrt{\lambda^2-2}+\sqrt{2} \\
\Rightarrow & \lambda^2+4=\lambda^2-2+2+2 \sqrt{2} \sqrt{\lambda^2-2} \\
\Rightarrow & 4=2 \sqrt{2} \sqrt{\lambda^2-2} \Rightarrow \sqrt{2}=\sqrt{\lambda^2-2} \\
\Rightarrow & 2=\lambda^2-2 \Rightarrow \lambda^2=4 \Rightarrow \lambda= \pm 2
\end{array}
$$
$$
\begin{array}{ll}
C_1 C_2=\text { sum of radii } \\
\Rightarrow & \sqrt{\lambda^2+4}=\sqrt{\lambda^2-2}+\sqrt{2} \\
\Rightarrow & \lambda^2+4=\lambda^2-2+2+2 \sqrt{2} \sqrt{\lambda^2-2} \\
\Rightarrow & 4=2 \sqrt{2} \sqrt{\lambda^2-2} \Rightarrow \sqrt{2}=\sqrt{\lambda^2-2} \\
\Rightarrow & 2=\lambda^2-2 \Rightarrow \lambda^2=4 \Rightarrow \lambda= \pm 2
\end{array}
$$
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