Chapter 6.4, Problem 42E

Solution

To convert: The polar equation to rectangular forms and identifies the graph, also support the answer by graphing the polar equation

The rectangular form of equation is ${\left(x-2\right)}^2+{\left(y+2\right)}^2=8$ and a circle with center at $\left(2,-2\right)$ and radius $2\sqrt{2}$ .

Given data:

The equation is $r=4\cos \theta -4\sin \theta$ .

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:

Multiply both sides of the equation by $r$ .

  $r^2=4r\cos \theta -4r\sin \theta$

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

  $\begin{array}{c}{x}^2+y^2=4x-4y\\ x^2+y^2-4x+4y=0\\ x^2-4x+4+y^2+4y+4=0+4+4\\ {\left(x-2\right)}^2+{\left(y+2\right)}^2=8\end{array}$

The graph is a circle with center at $\left(2,-2\right)$ and radius $2\sqrt{2}$ .

This is supported by using a graphing calculator in Polar and Radian mode as shown below.

  

Therefore, the rectangular form of equation is ${\left(x-2\right)}^2+{\left(y+2\right)}^2=8$ and a circle with center at $\left(2,-2\right)$ and radius $2\sqrt{2}$ .