Chapter 6.4, Problem 41E

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 ${\left(x+2\right)}^2+{\left(y-1\right)}^2=5$ and a circle with center at $\left(-2,1\right)$ and radius $\sqrt{5}$ .

Given data:

The equation is $r=2\sin \theta -4\cos \theta$ .

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:

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

  $\begin{array}{c}{x}^2+y^2=2y-4x\\ x^2+4x+y^2-2y=0\\ \left(x^2+4x+{\left(-\frac{4}{2}\right)}^2\right)+\left(y^2-2y+{\left(-\frac{2}{2}\right)}^2\right)=0+{\left(\frac{4}{2}\right)}^2+{\left(-\frac{2}{2}\right)}^2\\ {\left(x+2\right)}^2+{\left(y-1\right)}^2=5\end{array}$

The graph is a circle with center at $\left(-2,1\right)$ and radius $\sqrt{5}$ .

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

  

Therefore, the rectangular form of equation is ${\left(x+2\right)}^2+{\left(y-1\right)}^2=5$ and a circle with center at $\left(-2,1\right)$ and radius $\sqrt{5}$ .