Chapter 9.2, Problem 27AYU

In Problems 15-30, transform each polar equation to an equation in rectangular coordinates. Then identify and graph the equation.

$r\csc \theta =-2$

To determine

To find: Transform the polar equation $r \csc \theta = - 2$ to rectangular coordinates. Then identify and graph the equation.

Answer to Problem 27AYU

Solution:

The graph of $r \csc \theta = - 2$ is the equation of circle $x^2 + {\left(\text{y} + 1\right)}^2 = 1^2$ with center $\left(0, - 1\right)$ and radius 1.

Given:

It is asked to transform the polar equation $r \csc \theta = - 2$ to rectangular coordinates and graph it.

Explanation of Solution

Convert the polar equation to a rectangular equation $\csc \theta = - 2$ .

Formula used:

$\begin{array}{c}\left(x, y\right) = \left(r \text{cos}\theta , r \text{sin}\theta \right)\\ x^2 + y^2 = {\text{r}}^2\end{array}$

Consider $\csc \theta = - 2$ ,

$\begin{array}{c}r\left(\frac{1}{\text{sin}\theta }\right) = - 2\\ r = - 2 \text{sin}\theta \end{array}$

Multiply both the sides by $r$,

$\begin{array}{c}{r}^2 = - 2r \text{sin}\theta \\ x^2 + y^2 = - 2y\\ x^2 + y^2 + 2y = 0\end{array}$

Now use completion of square for ‘$y$’.

$\begin{array}{c}{x}^2 + y^2 + 2y + 1 = 1\\ x^2 + {(\text{y} + 1)}^2 = 1^2\end{array}$

The above is an equation of circle with center $\left(0, - 1\right)$ and radius 1.

The graph of $r \csc \theta = - 2$ is the equation of circle $x^2 + {\left(\text{y} + 1\right)}^2 = 1^2$ with center $\left(0, - 1\right)$ and radius 1.

To sketch:

$r \csc \theta = - 2$