If the circles x²+y²=2 and x²+y²–4x–4y+$\mathrm{\lambda }$=0 have exactly three real common tangents the $\mathrm{\lambda }$=

$$\begin{gathered} C_1=\left(0,0\right),r_1=\sqrt{2},C_2=(2,2),r_2=\sqrt{8-\lambda } \\ \mathrm{since} \mathrm{the} \mathrm{circles} \mathrm{have} \mathrm{exactlly} \mathrm{three} \mathrm{common} \mathrm{tangents} \\ {C}_1{C}_2=r_1+r_2\Rightarrow \sqrt{8}=\sqrt{2}+\sqrt{8-\lambda } \\ \sqrt{8-\lambda }=\sqrt{2} \\ \therefore \lambda =6 \end{gathered}$$