Chapter 10, Problem 9CR

To determine

The polar equation for the circle with center at the point $\left(0,4\right)$ and radius $4$ and graph the circle.

Answer to Problem 9CR

Solution:

The polar equation of the circle with center at the point $\left(0,4\right)$ and radius $4$ is $r=8\sin \theta$ and the graph of the circle is

  

Explanation of Solution

Given information:

The circle has the center at the point $\left(0,4\right)$ and radius $4$

The general equation of the circle having center at $\left(h,k\right)$ and radius $r$ is ${\left(x-h\right)}^2+{\left(y-k\right)}^2=r^2$

Form the given information, the center of the given circle is $\left(h,k\right)=\left(0,4\right)$ and radius is $r=4$ .

Then, the equation of the given circle by substituting the above values is

  ${\left(x-0\right)}^2+{\left(y-4\right)}^2=4^2$

  $\Rightarrow x^2+{\left(y-4\right)}^2=16$

  $\Rightarrow x^2+y^2+16-8y=16$

  $\Rightarrow x^2+y^2-8y=0$

To convert the rectangular equation into polar equation, substitute polar coordinates $x=r\cos \theta ,y=r\sin \theta$

  $\Rightarrow {\left(r\cos \theta \right)}^2+{\left(r\sin \theta \right)}^2-8\left(r\sin \theta \right)=0$

  $\Rightarrow r^2{\cos }^2\theta +r^2{\sin }^2\theta -8r\sin \theta =0$

  $\Rightarrow r^2\left({\cos }^2\theta +{\sin }^2\theta \right)-8r\sin \theta =0$

Using the trigonometric identity ${\cos }^2\theta +{\sin }^2\theta =1$

  $\Rightarrow r^2-8r\sin \theta =0$

  $\Rightarrow r^2=8r\sin \theta$

  $\Rightarrow r=8\sin \theta$

The graph of the circle is as follows:

  

Therefore, the polar equation of the circle with center at the point $\left(0,4\right)$ and radius $4$ is $r=8\sin \theta$ .