Chapter 6.4, Problem 49E

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

Given data:

The equation is ${\left(x+3\right)}^2+{\left(y+3\right)}^2=18$ .

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.

  $\begin{array}{c}\left(x^2+6x+9\right)+{\left(y^2+6y+9\right)}^2=18\\ \left(x^2+y^2\right)+6x+6y=0\end{array}$

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

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

The graph of the rectangular equation is a circle centered at $\left(-3,-3\right)$ and radius $3\sqrt{2}$ as shown below.

  

Figure (1)

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

  

Figure (2)

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