If a variable line has its intercepts on the coordinate axes e and e' , where e/2 and e'/ 2 are the eccentricities of a hyperbola and its conjugate hyperbola, then the line always touches the circle x² + y² = r² , where r =

Since $\frac{\mathrm{e}}{2} \mathrm{and} \frac{{\mathrm{e}}^{\text{'}}}{2}$ are the eccentricities of a hyperbola and its conjugate hyperbola, 
we have  $\frac{4}{{\mathrm{e}}^2}+\frac{4}{{\mathrm{e}}^{\mathrm{\prime }2}}=1\text{ or }4=\frac{{\mathrm{e}}^2{\mathrm{e}}^{\mathrm{\prime }2}}{{\mathrm{e}}^{\mathrm{\prime }2}+{\mathrm{e}}^{\mathrm{\prime }2}}$ 
The line passing through the points (e, 0) and (0,e') is ${\mathrm{e}}^{\mathrm{\prime }}\mathrm{x}+\mathrm{ey}-{\mathrm{ee}}^{\mathrm{\prime }}=0$ 
 It is tangent to the circle ${\mathrm{x}}^2+{\mathrm{y}}^2={\mathrm{r}}^2.$  
Hence, $\frac{{\mathrm{ee}}^{\mathrm{\prime }}}{\sqrt{{\mathrm{e}}^2+{\mathrm{e}}^{\mathrm{\prime }2}}}=\mathrm{r}\text{ or }{\mathrm{r}}^2=\frac{{\mathrm{e}}^2{\mathrm{e}}^{\mathrm{\prime }2}}{{\mathrm{e}}^2+{\mathrm{e}}^{\mathrm{\prime }2}}=4\text{ or }\mathrm{r}=2$