The goal is to eliminate [latex]x[/latex] and [latex]y[/latex] from the equation and introduce [latex]r[/latex] and [latex]\theta[/latex]. Ideally, we would write the equation [latex]r[/latex] as a function of [latex]\theta[/latex]. To obtain the polar form, we will use the relationships between [latex]\left(x,y\right)[/latex] and [latex]\left(r,\theta \right)[/latex]. Since [latex]x=r\cos \theta[/latex] and [latex]y=r\sin \theta[/latex], we can substitute and solve for [latex]r[/latex].

[latex]\begin{align}&{\left(r\cos \theta \right)}^{2}+{\left(r\sin \theta \right)}^{2}=9 \\ &{r}^{2}{\cos }^{2}\theta +{r}^{2}{\sin }^{2}\theta =9 \\ &{r}^{2}\left({\cos }^{2}\theta +{\sin }^{2}\theta \right)=9 \\ &{r}^{2}\left(1\right)=9&& {\text{Substitute cos}}^{2}\theta +{\sin }^{2}\theta =1. \\ &r=\pm 3&& \text{Use the square root property}. \end{align}[/latex]

Thus, [latex]{x}^{2}+{y}^{2}=9,r=3[/latex], and [latex]r=-3[/latex] should generate the same graph.To graph a circle in rectangular form, we must first solve for [latex]y[/latex].

[latex]\begin{gathered} {x}^{2}+{y}^{2}=9 \\ {y}^{2}=9-{x}^{2} \\ y=\pm \sqrt{9-{x}^{2}}\end{gathered}[/latex]

Note that this is two separate functions, since a circle fails the vertical line test. Therefore, we need to enter the positive and negative square roots into the calculator separately, as two equations in the form [latex]{Y}_{1}=\sqrt{9-{x}^{2}}[/latex] and [latex]{Y}_{2}=-\sqrt{9-{x}^{2}}[/latex]. Press GRAPH.

This equation appears similar to the previous example, but it requires different steps to convert the equation.We can still follow the same procedures we have already learned and make the following substitutions:

[latex]\begin{align}&{r}^{2}=6y&& \text{Use }{x}^{2}+{y}^{2}={r}^{2}. \\ &{r}^{2}=6r\sin \theta&& \text{Substitute}y=r\sin \theta . \\ &{r}^{2}-6r\sin \theta =0&& \text{Set equal to 0}. \\ &r\left(r - 6\sin \theta \right)=0&& \text{Factor and solve}. \\ &r=0&& \text{We reject }r=0,\text{as it only represents one point, }\left(0,0\right). \\ &\text{or }r=6\sin \theta \end{align}[/latex]

Therefore, the equations [latex]{x}^{2}+{y}^{2}=6y[/latex] and [latex]r=6\sin \theta[/latex] should give us the same graph.

The Cartesian or rectangular equation is plotted on the rectangular grid, and the polar equation is plotted on the polar grid. Clearly, the graphs are identical.

We will use the relationships [latex]x=r\cos \theta[/latex] and [latex]y=r\sin \theta[/latex].

[latex]\begin{align}&y=3x+2 \\ &r\sin \theta =3r\cos \theta +2 \\ &r\sin \theta -3r\cos \theta =2 \\ &r\left(\sin \theta -3\cos \theta \right)=2&& \text{Isolate }r. \\ &r=\frac{2}{\sin \theta -3\cos \theta }&& \text{Solve for }r. \end{align}[/latex]