# To find: The equivalent polar equation for the given rectangular-coordinate equation.Given:\displaystyle{x}^{2}+{y}^{2}+{8}{x}={0}

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

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Concept used: Conversion formula for coordinate systems are given as
a)
From polar to rectangular:
$x=r\mathrm{cos}\theta$
$y=r\mathrm{sin}\theta$
b)
From rectangular to polar:
$r=±\sqrt{{x}^{2}+{y}^{2}}$
$\mathrm{cos}\theta =\frac{x}{r};\mathrm{sin}\theta =\frac{y}{r};\mathrm{tan}\theta =\frac{x}{y}$
Calculation:
Converting into equivalent polar equation
${x}^{2}+{y}^{2}+8x=0$
Put $\left\{x\right\}=\left\{r\right\}\mathrm{cos}\left\{\theta \right\},\left\{y\right\}=\left\{r\right\}\mathrm{sin}\left\{\theta \right\},$
$x=r\mathrm{cos}\theta ,y=r\mathrm{sin}\theta$
$⇒{\left(r\mathrm{cos}\theta \right)}^{2}+{\left(r\mathrm{sin}\theta \right)}^{2}+8r\mathrm{cos}\theta =0$
$⇒{r}^{2}{\mathrm{cos}}^{2}\theta +{r}^{2}{\mathrm{sin}}^{2}\theta +8r\mathrm{cos}\theta =0$
$⇒{r}^{2}\left({\mathrm{cos}}^{2}\theta +{\mathrm{sin}}^{2}\theta \right)+8r\mathrm{cos}\theta =0\left\{{\mathrm{cos}}^{2}\theta +{\mathrm{sin}}^{2}\theta =1\right\}$
$⇒{r}^{2}+8r\mathrm{cos}\theta =0$
$⇒r+8\mathrm{cos}\theta =0$
Thus, desired equivalent polar equation would be $r+8\mathrm{cos}\theta =0$