The equivalent polar equation for the given rectangular-coordinate equation: x^{2} + y^{2} + 8x=0

Conversion formula for coordinate systems are given as:
a) From polar to rectangular:
${\displaystyle x=r\cos \theta }$
${\displaystyle y=r\sin \theta }$
b) From rectangular to polar:
${\displaystyle r=\pm \sqrt{x^2+y^2}}$
${\displaystyle \cos \theta =\frac{x}{r},\sin \theta =\frac{y}{r},\tan \theta =\frac{x}{y}}$
Converting into equivalent polar coordinates:
${\displaystyle [x^2+y^2+8x=0}$
Put ${\displaystyle x=r\cos \theta ,y=r\sin \theta ,}$
${\displaystyle \Rightarrow {\left(r\cos \theta \right)}^2+{\left(r\sin \theta \right)}^2+8r\cos \theta =0}$
${\displaystyle \Rightarrow r^2{\cos }^2\theta +r^2{\sin }^2\theta +8r\cos \theta =0}$
${\displaystyle \Rightarrow r^2({\cos }^2\theta +{\sin }^2\theta )+8r\cos \theta =0}$
${\displaystyle \{{\cos }^2\theta +{\sin }^2\theta =0\}}$
${\displaystyle \Rightarrow r^2+8r\cos \theta =0}$
${\displaystyle \Rightarrow r+8\cos \theta =0}$
Hence, desired equivalent polar coordinates would be ${\displaystyle r+8\cos \theta =0}$