Locus of feet of perpendicular from (5, 0) to the tangents of $\frac{{\mathrm{x}}^2}{16}-\frac{{\mathrm{y}}^2}{9}=1$ is

$$\begin{gathered} Given hyperbola \frac{{\mathrm{x}}^2}{16}-\frac{{\mathrm{y}}^2}{9}=1 \\ where {\mathrm{a}}^2=16, {\mathrm{b}}^2=9 \\ \mathrm{Now} \mathrm{e}=\sqrt{\frac{16+9}{16}}=\frac{5}{4} \\ \mathrm{Now} \mathrm{focus}=(\mathrm{ae}, 0) \\ =(5, 0) \mathrm{which} \mathrm{is} \mathrm{the} \mathrm{given} \mathrm{point} \\ \therefore \mathrm{The} \mathrm{locus} \mathrm{of} \mathrm{foot} \mathrm{of} \mathrm{perpendicular} \mathrm{from} \mathrm{foci} \mathrm{to} \mathrm{any} \mathrm{tangent} \mathrm{to} \mathrm{hyperbola} \mathrm{is} \mathrm{auxilary} \mathrm{circle} \\ \mathrm{Required} \mathrm{locus} \mathrm{is} {\mathrm{x}}^2+{\mathrm{y}}^2=a^2 \Rightarrow {\mathrm{x}}^2+{\mathrm{y}}^2=16 \end{gathered}$$