$$\begin{gathered} \mathrm{The} \mathrm{circle} {\mathrm{x}}^2+{\mathrm{y}}^2-8\mathrm{x}=0 \mathrm{and} \mathrm{hyperbola} \frac{{\mathrm{x}}^2}{9}-\frac{{\mathrm{y}}^2}{4}=1 \mathrm{intersect} \mathrm{at} \mathrm{the} \mathrm{points} \mathrm{A} \mathrm{and} \mathrm{B}. \mathrm{Equation} \mathrm{of} \mathrm{a} \mathrm{common} \\ \mathrm{tangent} \mathrm{with} \mathrm{positive} \mathrm{slope} \mathrm{to} \mathrm{the} \mathrm{circle} \mathrm{as} \mathrm{well} \mathrm{as} \mathrm{to} \mathrm{the} \mathrm{hyperbola} \mathrm{is} \end{gathered}$$

$$\begin{gathered} Given hyperbola \frac{{\mathrm{x}}^2}{9}-\frac{{\mathrm{y}}^2}{4}=1....\left(1\right) where a^2=9,b^2=4 \\ \text{ Any tangent to hyperbola is }\mathrm{y}=\mathrm{mx}+\sqrt{9{\mathrm{m}}^2-4}\text{, →(1)} \\ Given \text{ circle }{\mathrm{x}}^2+{\mathrm{y}}^2-8\mathrm{x}=0 \\ Equation of \tan gent in slope form is y=m(x-4)+\sqrt{16m^2+16}\to \left(2\right) \\ \mathrm{Since} (1 ) \mathrm{and} \left(2\right) \mathrm{are} \mathrm{same} \mathrm{then} \mathrm{m}=\frac{2}{\sqrt{5}} \\ \therefore \mathrm{common} \mathrm{tangent} \mathrm{is} \mathrm{sub} \mathrm{m} \mathrm{in} \left(1\right) \\ \mathrm{y}=\frac{2}{\sqrt{5}}\mathrm{x}+\sqrt{9\left(\frac{4}{5}\right)-4} \\ \Rightarrow \mathrm{y}=\frac{2}{\sqrt{5}}\mathrm{x}+\frac{4}{\sqrt{5}} \\ \Rightarrow 2\mathrm{x}-\sqrt{5}\mathrm{y}+4=0 \end{gathered}$$