Question

How do you convert $r=4$ into Cartesian form? \[\]

Answer 1

Hint: We recall polar and Cartesian representations of a point in the plane. We use the conversion from rule from polar coordinates $\left( r,\theta \right)$ to $\left( x,y \right)=\left( r\cos \theta ,r\sin \theta \right)$ and eliminate $\theta $ to have ${{r}^{2}}={{x}^{2}}+{{y}^{2}}$. We put $r=4$ in the obtained equation and simplify.\[\]

Complete step by step answer:
We know that in the Cartesian coordinate system two perpendicular intersecting lines are taken as reference axes called $x-$axis and $y-$axis. Any point on the plane is denoted as $\left( x,y \right)$ where $x$ is the distance from $y-$axis and $y$is the distance from $x-$axis. \[\]
We know in the polar coordinate system a ray and a point are taken as reference. The ray is called the polar axis and the point is called pole. Any point on the plane at a distance be $r$ and angle subtended ray joining from pole to that point with the polar axis be $\theta $ and the polar coordinates of that point $\left( r,\theta \right)$ where $r$ is called radial coordinate and $\theta $is called angular coordinate.\[\]
If we take positive $x-$axis as the polar axis and origin as the pole then any coordinate any in polar form $\left( r,\theta \right)$ can be converted to Cartesian form $\left( x,y \right)$ using working rule
\[\begin{align}
  & x=r\cos \theta \\
 & y=r\sin \theta \\
\end{align}\]
 We square each of the above equation to have
\[\begin{align}
  & {{x}^{2}}={{r}^{2}}{{\cos }^{2}}\theta \\
 & {{y}^{2}}={{r}^{2}}{{\sin }^{2}}\theta \\
\end{align}\]
We add respective sides of each equation in order to eliminate $\theta $. We have
 \[\begin{align}
  & {{x}^{2}}+{{y}^{2}}={{r}^{2}}{{\cos }^{2}}\theta +{{r}^{2}}{{\sin }^{2}}\theta \\
 & \Rightarrow {{x}^{2}}+{{y}^{2}}={{r}^{2}}\left( {{\cos }^{2}}\theta +{{\sin }^{2}}\theta \right) \\
\end{align}\]
We use the Pythagorean trigonometric identity ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$ in the above step to have;
\[\Rightarrow {{x}^{2}}+{{y}^{2}}={{r}^{2}}\]
We are given that $r=4$. We put $r=4$ in the above step to have;
\[\begin{align}
  & \Rightarrow {{x}^{2}}+{{y}^{2}}={{4}^{2}} \\
 & \Rightarrow {{x}^{2}}+{{y}^{2}}=16 \\
\end{align}\]
The above equation represents $\left( x,y \right)$ such that $r=4$. \[\]

Note: We see that the locus of all the points with $r=4$ in polar form is circle with centre at origin and radius 4. The general equation of circle with centre $\left( a,b \right)$ and radius $r$ is given as ${{\left( x-a \right)}^{2}}+{{\left( y-b \right)}^{2}}={{r}^{2}}$ which in parametric form is $\left( a+r\cos \theta ,b+r\sin \theta \right)$. Here in this problem $\left( a,b \right)=\left( 0,0 \right)$ and all points on ${{x}^{2}}+{{y}^{2}}=16$ in parametric form are $\left( 4\cos \theta ,4\sin \theta \right),\theta \in \left[ 0,2\pi \right)$. We can convert from Cartesian $\left( x,y \right)$ form to polar form $\left( r,\theta \right)$ such that $r=\sqrt{{{x}^{2}}+{{y}^{2}}}$ and $\theta ={{\tan }^{-1}}\left( \dfrac{y}{x} \right)$ with the condition on $\theta $ that $\theta $ will be measured in clockwise or anticlockwise sense from positive $x-$axis (polar axis) to minimize the amount of rotation.