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

Question
Alternate coordinate systems
The equivalent polar equation for the given rectangular-coordinate equation:
$$\displaystyle{x}^{{{2}}}\ +\ {y}^{{{2}}}\ +\ {8}{x}={0}$$

2020-12-25
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:
$$\displaystyle{\left[{x}^{{{2}}}\ +\ {y}^{{{2}}}\ +\ {8}{x}={0}\right.}$$
Put $$\displaystyle{x}={r}{\cos{\theta}},\ {y}={r}{\sin{\theta}},$$
$$\displaystyle\Rightarrow\ {\left({r}{\cos{\theta}}\right)}^{{{2}}}\ +\ {\left({r}{\sin{\theta}}\right)}^{{{2}}}\ +\ {8}{r}{\cos{\theta}}={0}$$
$$\displaystyle\Rightarrow\ {r}^{{{2}}}{{\cos}^{{{2}}}\theta}\ +\ {r}^{{{2}}}{{\sin}^{{{2}}}\theta}\ +\ {8}{r}{\cos{\theta}}={0}$$
$$\displaystyle\Rightarrow\ {r}^{{{2}}}\ {\left({{\cos}^{{{2}}}\theta}\ +{{\sin}^{{{2}}}\ }\theta\right)}\ +\ {8}{r}{\cos{\theta}}={0}$$
$$\displaystyle\le{f}{t}{\left\lbrace{{\cos}^{{{2}}}\theta}\ +{{\sin}^{{{2}}}\ }\theta={0}{r}{i}{g}{h}{t}\right\rbrace}$$
$$\displaystyle\Rightarrow\ {r}^{{{2}}}\ +\ {8}{r}{\cos{\theta}}={0}$$
$$\displaystyle\Rightarrow\ {r}\ +\ {8}{\cos{\theta}}={0}$$
Hence, desired equivalent polar coordinates would be $$\displaystyle{r}\ +\ {8}{\cos{\theta}}={0}$$

### Relevant Questions

To find: The equivalent polar equation for the given rectangular-coordinate equation.
Given:
$$\displaystyle{d}{i}{s}{p}{l}{a}{y}{s}{t}{y}\le{\left\lbrace{x}\right\rbrace}^{{{2}}}+{\left\lbrace{y}\right\rbrace}^{{{2}}}+{\left\lbrace{8}\right\rbrace}{\left\lbrace{x}\right\rbrace}={\left\lbrace{0}\right\rbrace}$$
To find: The equivalent polar equation for the given rectangular-coordinate equation.
Given:
$$\displaystyle\ {x}=\ {r}{\cos{\theta}}$$
$$\displaystyle\ {y}=\ {r}{\sin{\theta}}$$
b. From rectangular to polar:
$$\displaystyle{r}=\pm\sqrt{{{x}^{{{2}}}\ +\ {y}^{{{2}}}}}$$
$$\displaystyle{\cos{\theta}}={\frac{{{x}}}{{{r}}}},{\sin{\theta}}={\frac{{{y}}}{{{r}}}},{\tan{\theta}}={\frac{{{x}}}{{{y}}}}$$
Calculation:
Given: equation in rectangular-coordinate is $$\displaystyle{y}={x}$$.
Converting into equivalent polar equation -
$$\displaystyle{y}={x}$$
Put $$\displaystyle{x}={r}{\cos{\theta}},\ {y}={r}{\sin{\theta}},$$
$$\displaystyle\Rightarrow\ {r}{\sin{\theta}}={r}{\cos{\theta}}$$
$$\displaystyle\Rightarrow\ {\frac{{{\sin{\theta}}}}{{{\cos{\theta}}}}}={1}$$
$$\displaystyle\Rightarrow\ {\tan{\theta}}={1}$$
Thus, desired equivalent polar equation would be $$\displaystyle\theta={1}$$
The equivalent polar equation for the given rectangular - coordinate equation.
$$\displaystyle{y}=\ -{3}$$
The equivalent polar coordinates for the given rectangular coordinates.
A rectangular coordinate is given as (0, -3).
Find the Equivalent Polar Equation for a given Equation with Rectangular Coordinates:
$$\displaystyle{r}{\cos{\theta}}=\ -{1}$$
The equivalent polar coordinates for the given rectangular coordinates:
$$\displaystyle{\left[\begin{array}{cc} {4}&\ {0}\end{array}\right]}$$
$$\displaystyle{\left[\begin{array}{cc} {6}&\ -{135}^{{\circ}}\end{array}\right]}$$
The equivalent polar coordinates for the given rectangular coordinates: $$\displaystyle{\left[\begin{array}{cc} {5}&\ {180}^{{\circ}}\end{array}\right]}$$
Interraption: To show that the system $$\displaystyle{d}{i}{s}{p}{l}{a}{y}{s}{t}{y}\le\dot{{{\left\lbrace{r}\right\rbrace}}}={\left\lbrace{r}\right\rbrace}{\left\lbrace{\left({\left\lbrace{1}\right\rbrace}-{\left\lbrace{r}\right\rbrace}^{{{2}}}\right)}\right\rbrace},\dot{{\theta}}={\left\lbrace{1}\right\rbrace}$$ is equivalent to $$\displaystyle{d}{i}{s}{p}{l}{a}{y}{s}{t}{y}\le\dot{{{\left\lbrace{x}\right\rbrace}}}={\left\lbrace{x}\right\rbrace}-{\left\lbrace{y}\right\rbrace}-{\left\lbrace{x}\right\rbrace}{\left\lbrace{\left({\left\lbrace{x}\right\rbrace}^{{{2}}}+{\left\lbrace{y}\right\rbrace}^{{{2}}}\right)}\right\rbrace},\dot{{{\left\lbrace{y}\right\rbrace}}}={\left\lbrace{x}\right\rbrace}+{\left\lbrace{y}\right\rbrace}-{\left\lbrace{y}\right\rbrace}{\left\lbrace{\left({\left\lbrace{x}\right\rbrace}^{{{2}}}+{\left\lbrace{y}\right\rbrace}^{{{2}}}\right)}\right\rbrace}$$ for polar to Cartesian coordinates.