Chapter 4.4, Problem 75E

(a)

To determine

To rewrite: given equation of a circle in standard form, to identify the center and radius of the circle.

Answer to Problem 75E

The center of circle is $\left(-3,0\right)$ and radius is 5.

Explanation of Solution

Given information:

The standard equation of a circle with radius $r$ and center $\left(h,k\right)$ is ${\left(x-h\right)}^2+{\left(y-k\right)}^2=r^2$ .

Given graph,

  

Given equation,

  $x^2+6x+9+y^2=25$

Calculation:

Consider the equation $x^2+6x+9+y^2=25$ :

Rewrite the equation as

  $\begin{array}{c}\left(x^2+6x+9\right)+y^2=25\\ {\left(x+3\right)}^2+y^2=5^2\end{array}$

Comparing with ${\left(x-h\right)}^2+{\left(y-k\right)}^2=r^2$ , have $h=-3,k=0$ and $r=5$ .

Therefore, the center of circle is $\left(-3,0\right)$ and radius is 5.

The graph of circle is shown below:

  

(b)

To determine

To rewrite: given equation of a circle in standard form, to identify the center and radius of the circle.

Answer to Problem 75E

The center of circle is $\left(2,0\right)$ and radius is 3.

Explanation of Solution

Given information:

The standard equation of a circle with radius $r$ and center $\left(h,k\right)$ is ${\left(x-h\right)}^2+{\left(y-k\right)}^2=r^2$ .

Given graph,

  

Given equation,

  $x^2-4x+4+y^2=9$

Calculation:

Consider the equation $x^2-4x+4+y^2=9$ :

Rewrite the equation as

  $\begin{array}{c}{x}^2-4x+4+y^2=9\\ {\left(x-2\right)}^2+y^2=3^2\end{array}$

Comparing with ${\left(x-h\right)}^2+{\left(y-k\right)}^2=r^2$ , have $h=2,k=0$ and $r=3$ .

Therefore, the center of circle is $\left(2,0\right)$ and radius is 3.

The graph of circle is shown below:

  

(c)

To determine

To rewrite: given equation of a circle in standard form, to identify the center and radius of the circle.

Answer to Problem 75E

The center of circle is $\left(4,-1\right)$ and radius is 6.

Explanation of Solution

Given information:

The standard equation of a circle with radius $r$ and center $\left(h,k\right)$ is ${\left(x-h\right)}^2+{\left(y-k\right)}^2=r^2$ .

Given graph,

  

Given equation,

  $x^2-8x+16+y^2+2y+1=36$

Calculation:

Consider the equation $x^2-8x+16+y^2+2y+1=36$ :

Rewrite the equation as

  $\begin{array}{c}{x}^2-8x+16+y^2+2y+1=36\\ {\left(x-4\right)}^2+{\left(y+1\right)}^2=6^2\end{array}$

Comparing with ${\left(x-h\right)}^2+{\left(y-k\right)}^2=r^2$ , have $h=4,k=-1$ and $r=6$ .

Therefore, the center of circle is $\left(4,-1\right)$ and radius is 6.

The graph of circle is shown below: