Chapter 13.1, Problem 37WE

To determine

To find the equation of the circle and sketch the graph

Answer to Problem 37WE

The equation of the circle are $x^2+{\left(y-2\right)}^2=9$

Explanation of Solution

Given information:

The circle has diameter RS where $R\text{ is }\left(-3,2\right)\text{ and S is }\left(3,2\right)$

Calculation:

The equation of the circle can be obtained as

The standard form of the equation of a circle with center 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

First find the radius

  $d=RS=2r\left[\begin{array}{l}d=\text{diameter}\\ r=\text{radius}\end{array}\right]$

Therefore,

  $\begin{array}{l}RS=\sqrt{{\left(3-\left(-3\right)\right)}^2+{\left(2-2\right)}^2}\\ RS=\sqrt{{\left(3+3\right)}^2+{\left(0\right)}^2}\\ RS=\sqrt{{\left(6\right)}^2}\\ RS=6\end{array}$

  $\begin{array}{l}2r=RS\\ 2r=6\\ r=3\left[\text{divide both side by 2}\right]\end{array}$

Find the center

  $\begin{array}{l}c=\frac{R+S}{2}\left[c=\text{center}\right]\\ c=\frac{-3+3}{2},\frac{2+2}{2}\\ c=\frac{0}{2},\frac{4}{2}\\ c=\left(0,2\right)\end{array}$

Again put the values of $a=0,b=2\text{ and }r=3$ on the standard form of the equation

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

Hence,

The equation of the circle are $x^2+{\left(y-2\right)}^2=9$

Graph : Sketch the graph using graphing utility.

Step 1: Press WINDOW button to access the Window editor.

Step 2: Press $\boxed{\text{Y=}}$ button.

Step 3: Enter the expression $x^2+{\left(y-2\right)}^2=9$ which is required to graph.

Step 4: Press GRAPH button to graph the function and adjust the windows according to the graph.

The graph is obtained as:

  

Interpretation :

From the above graph, it can be observed that the center of a circle is $\left(0,2\right)$ and radius is 3.