Chapter 13.1, Problem 21WE

To determine

To find the equation of the circle

Answer to Problem 21WE

The equation of the circle are ${\left(x-3\right)}^2+y^2=64$

Explanation of Solution

The given center and radius is

  $C\left(3,0\right);r=8$

Calculation:

The equation of the circle can be obtained as

The standard form of the equation of a circle with center and radius is

  ${\left(x-a\right)}^2+{\left(y-b\right)}^2=r^2$

Where $\left(a,b\right)$ is the center and r is the radius

So, the center $\left(a,b\right)=\left(3,0\right)\text{ and radius }\left(r\right)=\left(8\right)$

Put the values of $a=3,b=0\text{ and }r=8$ on the standard form of the equation

  $\begin{array}{l}{\left(x-3\right)}^2+{\left(y-0\right)}^2={\left(8\right)}^2\\ {\left(x-3\right)}^2+y^2=64\end{array}$

Hence,

The equation of the circle are ${\left(x-3\right)}^2+y^2=64$