4.0 Relation between cartesian co-ordinates and polar co-ordinates

Let $P(x,y)$ be the cartesian co-ordinates with respect to axes $OX$ and $OY$ and $(r,\theta )$ be its polar co-ordinates with respect to pole $O$ and initial line $OX$ as shown in figure. $$\begin{equation} \begin{aligned} OM = x = r\cos \theta ...(1) \\ MP = y = r\sin \theta ...(2) \\\end{aligned} \end{equation} $$

Squaring and adding $(1)$ and $(2)$, we get $$\begin{equation} \begin{aligned} {x^2} + {y^2} = {r^2}({\cos ^2}\theta + {\sin ^2}\theta ) = {r^2} \\ r = \sqrt {{x^2} + {y^2}} \\\end{aligned} \end{equation} $$

Dividing $(2)$ by $(1)$, we get $$\begin{equation} \begin{aligned} \tan \theta = \frac{y}{x} \\ \theta = {\tan ^{ - 1}}\frac{y}{x} \\\end{aligned} \end{equation} $$

Therefore, we can write $$\begin{equation} \begin{aligned} (r\cos \theta ,r\sin \theta ) \Rightarrow (x,y)...(3) \\ (\sqrt {{x^2} + {y^2}} ,{\tan ^{ - 1}}\frac{y}{x}) \Rightarrow (r,\theta )...(4) \\\end{aligned} \end{equation} $$

If $r$ and $\theta $ are known, we can find $(x,y)$ from equation $(3)$ and

If $x$ and $y$ are known, we can find $(r,\theta )$ from equation $(4)$.