If a variable line has its intercepts on the coordinate axes e, e', where $\frac{\mathrm{e}}{2},\frac{\mathrm{e}\text{'}}{2}$ 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=

Given e and e' are intercepts of a line 

Now equation of line is $\frac{{\displaystyle \mathrm{x}}}{{\displaystyle \mathrm{e}}}+\frac{{\displaystyle \mathrm{y}}}{{\displaystyle {\mathrm{e}}^1}}=1$  ….(1)

given that $\frac{e}{2} and \frac{e\text{'}}{2} are eccentricities of hyperbola and it\text{'}s conjugate hyperbola then$

$\begin{array}{l}\frac{\mathrm{e}}{2}=\sqrt{\frac{{\mathrm{a}}^2+{\mathrm{b}}^2}{{\mathrm{a}}^2}},\frac{{\mathrm{e}}^1}{2}=\sqrt{\frac{{\mathrm{a}}^2+{\mathrm{b}}^2}{{\mathrm{b}}^2}}\\ \Rightarrow \frac{1}{{\mathrm{e}}^2}+\frac{1}{{\mathrm{e}}_1^2}=\frac{1}{4}\dots .\left(2\right)\end{array}$

$$\begin{gathered} Given circle x^2+y^2=r^2 \\ centre=(0,0) ,radius=r \end{gathered}$$

Since (1) touches the  circle then r=d

$$\begin{gathered} where d=perpendicular dis\tan ce from (0,0) to \left(1\right) \\ \Rightarrow d=\frac{\left|-1\right|}{\sqrt{{\displaystyle \frac{1}{e^2}}+{\displaystyle \frac{1}{e{\text{'}}^2}}}} \\ \Rightarrow d=\frac{1}{{\displaystyle \frac{1}{2}}}=2 \end{gathered}$$

 $\Rightarrow {\mathrm{r}}^2={\mathrm{d}}^2\Rightarrow {\mathrm{r}}^2=4\Rightarrow \mathrm{r}=2$