Chapter 6.4, Problem 47E

Solution

To convert: The rectangular equation to polar form, sketches the graph of the rectangular equation, and supports the answer by graphing the polar equation with grapher.

The polar form of equation is $r=6\cos \theta$ and the graph of rectangular equation is shown in figure (1).

Given data:

The equation is ${\left(x-3\right)}^2+y^2=9$ .

Formula used:

Use the following conversion relationships between a polar coordinate $\left(r,\theta \right)$ and its corresponding rectangular coordinate $\left(x,y\right)$ .

  $\begin{array}{c}x=r\cos \theta \\ y=r\sin \theta \\ r^2=x^2+y^2\\ \tan \theta =\frac{y}{x}\end{array}$

Calculation:

Simplify the given rectangular equation.

  $\begin{array}{c}\left(x^2-6x+9\right)+y^2=9\\ \left(x^2+y^2\right)-6x=0\end{array}$

Substitute $r^2$ for $x^2+y^2$ and $r\cos \theta$ for $x$ in the above equation.

  $\begin{array}{c}{r}^2-6r\cos \theta =0\\ r^2=6r\cos \theta \\ r=6\cos \theta \end{array}$

The graph of the rectangular equation is a circle centered at $\left(3,0\right)$ and radius $3$ as shown below.

  

Figure (1)

Draw the graph of the polar equation $r=6\cos \theta$ as follows.

  

Figure (2)

Therefore, the polar form of equation is $r=6\cos \theta$ and the graph of rectangular equation is shown in figure (1).