If a variable line has its intercepts on the coordinate axes e and e' , where $\frac{\mathrm{e}}{2}$ and $\frac{\mathrm{e}\text{'}}{2}$ are the eccentricities of a hyperbola and its conjugate hyperbola, then the line always touches the circle ${\mathrm{x}}^2+{\mathrm{y}}^2={\mathrm{r}}^2$, where r =

 Since $\frac{\mathrm{e}}{2}$ and e$\frac{\mathrm{e}\text{'}}{2}$ are the eccentricities of a hyperbola and
its conjugate hyperbola, we have
$\begin{array}{r}\frac{4}{{\mathrm{e}}^2}+\frac{4}{{\mathrm{e}}^{\mathrm{\prime }2}}=1\\ \mathrm{or} 4=\frac{{\mathrm{e}}^2{\mathrm{e}}^{\mathrm{\prime }2}}{{\mathrm{e}}^2+{\mathrm{e}}^{\mathrm{\prime }2}}\end{array}$
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,

$\begin{array}{l}\frac{{\mathrm{ee}}^{\mathrm{\prime }}}{\sqrt{{\mathrm{e}}^2+{\mathrm{e}}^{\mathrm{\prime }2}}}=\mathrm{r}\\ \mathrm{or} {\mathrm{r}}^2=\frac{{\mathrm{e}}^2{\mathrm{e}}^{\mathrm{\prime }2}}{{\mathrm{e}}^2+{\mathrm{e}}^{\mathrm{\prime }2}}=4\\ \mathrm{or} \mathrm{r}=2\end{array}$