Chapter 6, Problem 67RE

Solution

To find: The polar form of the given rectangular equation and graph the polar equation.

The polar form of given rectangular equation is $r=6\cos \theta -2\sin \theta$ and the graph of the polar equation is a circle.

Given data:

The rectangular equation is ${\left(x-3\right)}^2+{\left(y+1\right)}^2=10$ .

Formula used:

Let $\left(r,\theta \right)$ and $\left(x,y\right)$ are the polar and rectangular coordinates of the point P.

Then,

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

Calculation:

  ${\left(x-3\right)}^2+{\left(y+1\right)}^2=10$

Expand and simplify above expression.

  $\begin{array}{l}{x}^2+9-6x+y^2+1+2y=10\\ x^2+y^2+2y-6x=0\end{array}$

Substituting $r^2$ for $x^2+y^2$ , $r\sin \theta$ for $y$ and $r\cos \theta$ for $x$ .

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

Therefore, the polar form of given equation is $r=6\cos \theta -2\sin \theta$ .

Plot the graph of equation.

  

Figure (1)

From the graph, it can be observed that the graph of the equation $r=6\cos \theta -2\sin \theta$ is a circle.

Therefore, the polar form of given rectangular equation is $r=6\cos \theta -2\sin \theta$ and the graph of the polar equation is a circle.