Chapter 13.1, Problem 40WE

a.

To determine

To find the radius of the circle

Answer to Problem 40WE

The value of $r_1=\sqrt{2}$ and $r_2=\sqrt{32}$

Explanation of Solution

Given information:

The given equation is

  $x^2+y^2=2\text{ and }{\left(x-3\right)}^2+{\left(y-3\right)}^2=32$

Calculation:

The radius 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

Comparing the given equation to the standard form of the equation

The first equation is

  $r_1=\sqrt{2}$

And the second equation is

  $r_2=\sqrt{32}$

Hence,

The value of $r_1=\sqrt{2}$ and $r_2=\sqrt{32}$

b.

To determine

To find the distance between the centers of the circle

Answer to Problem 40WE

The value of ${\text{c}}_1{c}_2=3\sqrt{2}$

Explanation of Solution

Given information:

The given equation is

  $x^2+y^2=2\text{ and }{\left(x-3\right)}^2+{\left(y-3\right)}^2=32$

Calculation:

The distance between the centers 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

Comparing the given equation to the standard form of the equation

The first equation has center is

  $c_1=\left(0,0\right)$

And the second equation has center is

  $c_2=\left(3,3\right)$

Therefore,

  $\begin{array}{l}{c}_1{c}_2=\sqrt{{\left(3-0\right)}^2+{\left(3-0\right)}^2}\\ c_1{c}_2=\sqrt{{\left(3\right)}^2+{\left(3\right)}^2}\\ c_1{c}_2=\sqrt{9+9}\\ c_1{c}_2=\sqrt{18}\\ {\text{c}}_1{c}_2=3\sqrt{2}\end{array}$

Hence,

The value of ${\text{c}}_1{c}_2=3\sqrt{2}$

c.

To determine

To explain the circle must be externally tangent

Answer to Problem 40WE

The circle must be externally tangent because the distance between the centers of the circle is equal to difference of the radius

Explanation of Solution

Given information:

The given equation is

  $x^2+y^2=2\text{ and }{\left(x-3\right)}^2+{\left(y-3\right)}^2=32$

The circle must be externally tangent can be explain as

In previous part of the question we find the values of $r_{1,}{r}_2{\text{ and c}}_1{c}_2$

  $\begin{array}{l}{\text{c}}_1{c}_2=\left(r_2-r_1\right)\\ {\text{c}}_1{c}_2=\left(\sqrt{32}-\sqrt{2}\right)\\ {\text{c}}_1{c}_2=\left(4\sqrt{2}-\sqrt{2}\right)\\ {\text{c}}_1{c}_2=3\sqrt{2}\end{array}$

The circle must be externally tangent because the distance between the centers of the circle is equal to difference of the radius

Hence,

The circle must be externally tangent because the distance between the centers of the circle is equal to difference of the radius

d.

To determine

To sketch the graph of the circle

Explanation of Solution

Given information:

The given equation is

  $x^2+y^2=2\text{ and }{\left(x-3\right)}^2+{\left(y-3\right)}^2=32$

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 $x^2+y^2=2\text{ and }{\left(x-3\right)}^2+{\left(y-3\right)}^2=32$ 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 radius of the circle are $r_1=\sqrt{2}\text{ and }{r}_2=4\sqrt{2}$ and distance between the centers of the circle are ${\text{c}}_1{c}_2=3\sqrt{2}$ .