Answer
$$\left( {9,\pi } \right)$$ or $$\left( { - 9,2\pi } \right)$$
Work Step by Step
$$\eqalign{
& \left( { - 9,0} \right) \Rightarrow x = - 9{\text{ and }}y = 0 \cr
& {\text{Converting to cartesian coordinates }}\left( {r,\theta } \right).{\text{ Where}} \cr
& r = \sqrt {{x^2} + {y^2}} {\text{ and }}\theta = {\tan ^{ - 1}}\left( {\frac{y}{x}} \right) \cr
& {\text{We obtain}}{\text{,}} \cr
& r = \sqrt {{{\left( { - 9} \right)}^2} + {{\left( 0 \right)}^2}} = 9 \cr
& {\text{Coordinate }}x{\text{ }}\left( {{\text{positive}}} \right){\text{, Coordinate }}y\left( {\text{0}} \right) \cr
& \theta = {\tan ^{ - 1}}\left( {\frac{0}{{ - 9}}} \right) + \pi = \pi \cr
& {\text{The polar coordinates are:}} \cr
& \left( {r,\theta } \right) = \left( {9,\pi } \right) \cr
& or \cr
& \left( { - r,\theta + \pi } \right) = \left( { - 9,2\pi } \right) \cr} $$
