Chapter 6.4, Problem 40E

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 ${\left(x-\frac{3}{2}\right)}^2+y^2=\frac{9}{4}$ and a circle with center at $\left(\frac{3}{2},0\right)$ and radius $\frac{3}{2}$ .

Given data:

The equation is $r\sec \theta =3$ .

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:

Divide both sides by $\sec \theta$ .

  $\begin{array}{c}r=\frac{3}{\sec \theta }\\ r=3\cos \theta \end{array}$

Multiply both sides by $r$ .

  $r^2=3r\cos \theta$

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

  $\begin{array}{c}{x}^2+y^2=3x\\ x^2-3x+y^2=0\\ \left(x^2-3x+{\left(-\frac{3}{2}\right)}^2\right)+y^2=0+{\left(-\frac{3}{2}\right)}^2\\ {\left(x-\frac{3}{2}\right)}^2+y^2=\frac{9}{4}\end{array}$

The graph is a circle with center at $\left(\frac{3}{2},0\right)$ and radius $\frac{3}{2}$ .

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

  

Therefore, the rectangular form of equation is ${\left(x-\frac{3}{2}\right)}^2+y^2=\frac{9}{4}$ and a circle with center at $\left(\frac{3}{2},0\right)$ and radius $\frac{3}{2}$ .