Chapter 9.1, Problem 28E

To determine

To find the center and radius of a circle and verify the graph using graphing utility

Explanation of Solution

Given information:

The equation is $x^2-6x+y^2+6y+14=0$

Concept used:

The standard form of the equation of a circle is:

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

The point $\left(h,k\right)$ is the center of the circle and the positive number $r$ is the radius of the circle.

Graph:

Reduce the given equation to the standard form ${\left(x-h\right)}^2+{\left(y-k\right)}^2=r^2$ ,

  $\begin{array}{l}{x}^2-6x+y^2+6y+14=0\\ \left(x^2-6x+9\right)+\left(y^2+6y+9\right)=-14+9+9\left[\begin{array}{l}\text{9 and 9 are added to each side of the equation}\\ \text{when completing the squares}\text{.}\end{array}\right]\\ {\left(x-3\right)}^2+{\left(y+3\right)}^2=4\end{array}$

Now, compare ${\left(x-3\right)}^2+{\left(y+3\right)}^2={\left(2\right)}^2$ with ${\left(x-h\right)}^2+{\left(y-k\right)}^2=r^2$ ,

  $\left(h,k\right)=\left(-3,3\right)\text{ and }r=2$

Therefore,

The graph is a circle whose center is $\left(-3,3\right)$ and whose radius is $r=2$ .

Plot several points that are one unit from the center. So, the points $\left(-1,3\right),\left(-3,5\right),\left(-5,3\right)\text{ and }\left(-3,1\right)$ are convenient.

Draw a circle that passes through four points which is shown in the figure below:

  

To verify the above graph, use graphing calculator to graph the given equation follow the steps given below:

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

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

Step 3: Enter the expression $x^2-6x+y^2+6y+14=0$ 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:

  

Hence, the graph is verified.

Interpretation:

From the above graph, it is observed that the graph is a circle whose center is $\left(-3,3\right)$ and whose radius is $r=2$ .