Chapter 6, Problem 63RE

Solution

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

The rectangular form of given equation is ${\left(x+\frac{3}{2}\right)}^2+{\left(y+1\right)}^2=\frac{13}{4}$ and the graph of the equation is a circle with centre $\left(\frac{-3}{2},-1\right)$ and radius $\frac{\sqrt{13}}{2}$ .

Given data:

The polar equation is $r=-3\cos \theta -2\sin \theta$ .

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:

  $r=-3\cos \theta -2\sin \theta$

Multiply by $r$ on both sides.

  $r^2=-3r\cos \theta -2r\sin \theta$

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

  $\begin{array}{c}{x}^2+y^2=-3x-2y\\ x^2+y^2+3x+2y=0\\ x^2+3x+y^2+2y+1-1=0\\ x^2+3x+{\left(y+1\right)}^2-1=0\end{array}$

Further simplify the equation.

  $\begin{array}{c}{x}^2+3x+{\left(y+1\right)}^2-1=0\\ x^2+2\times \frac{3}{2}\times x+{\left(\frac{3}{2}\right)}^2-{\left(\frac{3}{2}\right)}^2+{\left(y+1\right)}^2-1=0\\ {\left(x+\frac{3}{2}\right)}^2+{\left(y+1\right)}^2=\frac{13}{4}\end{array}$

Therefore, the rectangular equation is ${\left(x+\frac{3}{2}\right)}^2+{\left(y+1\right)}^2=\frac{13}{4}$ .

Plot the graph of equation.

  

Figure (1)

From the graph, it can be observed that the graph of the equation ${\left(x+\frac{3}{2}\right)}^2+{\left(y+1\right)}^2=\frac{13}{4}$ is a circle with centre $\left(\frac{-3}{2},-1\right)$ and radius $\frac{\sqrt{13}}{2}$ .

Therefore, the rectangular form of given equation is ${\left(x+\frac{3}{2}\right)}^2+{\left(y+1\right)}^2=\frac{13}{4}$ and the graph of the equation is a circle with centre $\left(\frac{-3}{2},-1\right)$ and radius $\frac{\sqrt{13}}{2}$ .