Chapter 1, Problem 22RE

To determine

To find: the center and radius of each circle and graph the circle.

Answer to Problem 22RE

The circle has center $\left(-2,0\right)$ and the radius $r=3$ .

  

Explanation of Solution

Given:

  ${\left(x+2\right)}^2+y^2=9$

Calculation:

The standard form of an equation of a circle with radius r and canter $\left(h,k\right)$ is

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

Since the equation is in standard form of an equation of a circle. To graph the equation, first compare the given equation to the standard form of the equation of a circle. The comparison yields information about the circle

  $\begin{array}{l}{\left(x+2\right)}^2+y^2=9\\ {\left(x+2\right)}^2+{\left(y-0\right)}^2=3^2{\left(x-h\right)}^2+{\left(y-k\right)}^2=r^2\end{array}$

  $h=-2$ , $k=0$ and $r=3$ . The circle has center $\left(-2,0\right)$ and the radius $r=3$ .

Circle has center $\left(-2,0\right)$ and a radius of 3 units. To graph this circle, first plot the center $\left(-2,0\right)$ . Since the radius is 3, locate 3 points on the circle by plotting points 3 units to the left, to the right, up, and down from the center. These four points can then be used as guides to obtain the graph.

  

To find the x -intercepts, if any, let $y=0$ . Then

  $\begin{array}{l}{\left(x+2\right)}^2+y^2=9\\ {\left(x+2\right)}^2+0^2=9y=0\\ x+2=\pm 3\text{ Simplify}\\ x=1,-5\text{ Solve for }x\end{array}$

The x-intercepts are $\left(1,0\right),\left(-5,0\right)$ .

To find the y -intercepts, if any, let $x=0$ . Then

  $\begin{array}{l}{\left(x+2\right)}^2+y^2=9\\ {\left(0+2\right)}^2+y^2=9x=0\\ 4+y^2=9\text{ Simplify}\\ y=\pm \sqrt{5}\text{ Solve for }y\end{array}$

The y-intercepts are $\left(0,\pm \sqrt{5}\right)$ .

Conclusion:

Therefore, the circle has center $\left(-2,0\right)$ and the radius $r=3$ .