The locus of the point $\left(l, m\right)$. If the line $lx+my=1$ touches the circle $x^2+y^2=a^2$ is 

$$\begin{gathered} Given circle {\mathrm{x}}^2+{\mathrm{y}}^2={\mathrm{a}}^2 \\ \mathrm{centre}\left(\mathrm{C}\right)=(0,0),\mathrm{r}=\mathrm{a} \\ \mathrm{Given} \mathrm{line} \mathrm{lx}+\mathrm{my}-1=0\to \left(1\right) \\ \mathrm{eq}uation -\left(1\right) \mathrm{touches} \mathrm{the} \mathrm{circle} then r=d \\ \Rightarrow {\mathrm{r}}^2={\mathrm{d}}^2 \\ {\mathrm{a}}^2={\left(\frac{\left|-1\right|}{\sqrt{{\mathrm{l}}^2+{\mathrm{m}}^2}}\right)}^2 \\ Now locus of the point (l,m) is \\ \Rightarrow {\mathrm{a}}^2=\frac{1}{{\mathrm{x}}^2+{\mathrm{y}}^2} \\ \Rightarrow {\mathrm{x}}^2+{\mathrm{y}}^2=\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{${\mathrm{a}}^2$}\right. \\ \Rightarrow {\mathrm{a}}^2({\mathrm{x}}^2+{\mathrm{y}}^2)=1 \end{gathered}$$