The set of values of ‘ $\lambda$’ for which the real circles $x^2+y^2={\lambda }^2$ and $x^2+y^2-2\sqrt{2}x-2\sqrt{2}y-\lambda =0$ have exactly one common tangent is

$$\begin{gathered} \mathrm{For} \mathrm{given} \mathrm{circles} \\ {\mathrm{x}}^2+{\mathrm{y}}^2={\mathrm{\lambda }}^2 \\ {\mathrm{C}}_1=(0, 0) , {\mathrm{r}}_1=\mathrm{\lambda } \\ {\mathrm{x}}^2+{\mathrm{y}}^2-2\sqrt{2\mathrm{x}}-2\sqrt{2\mathrm{y}}-\mathrm{\lambda }=0 \\ {\mathrm{C}}_2=\left(\sqrt{2}, \sqrt{2}\right) \\ and {\mathrm{r}}_2=\sqrt{4+\mathrm{\lambda }} \\ From given data {\mathrm{C}}_1 {\mathrm{C}}_2=\left|{\mathrm{r}}_1-{\mathrm{r}}_2\right| \\ \sqrt{4}=\left|\mathrm{\lambda }-\sqrt{4+\mathrm{\lambda }}\right| \end{gathered}$$

Squaring on both sides

$$\begin{gathered} 4={\mathrm{\lambda }}^2+4+\mathrm{\lambda }-2\mathrm{\lambda }\sqrt{4+\mathrm{\lambda }} \\ 2\mathrm{\lambda }\sqrt{4+\mathrm{\lambda }}={\mathrm{\lambda }}^2+\mathrm{\lambda } \\ 4{\mathrm{\lambda }}^2(4+\mathrm{\lambda })={\mathrm{\lambda }}^4+{\mathrm{\lambda }}^{\mathbf{2}}+2{\mathrm{\lambda }}^3 \\ 16{\mathrm{\lambda }}^2+4{\mathrm{\lambda }}^3={\mathrm{\lambda }}^4+{\mathrm{\lambda }}^2+2{\mathrm{\lambda }}^3 \\ {\mathrm{\lambda }}^4-15{\mathrm{\lambda }}^2-2{\mathrm{\lambda }}^3=0 \\ {\mathrm{\lambda }}^2\left({\mathrm{\lambda }}^2-2\mathrm{\lambda }-15\right)=0 \\ \lambda =0, \left(\mathrm{\lambda }-5\right) \left(\mathrm{\lambda }+3\right)=0 \\ \mathrm{\lambda }=\left\{0,5\right\}(\because \lambda is radius it should be positive) \end{gathered}$$