Chapter 1.9, Problem 28E

To determine

To graph: The circle ${\left(y-1\right)}^2+x^2=1$ by solve for $y$ .

Explanation of Solution

Given information:

The equation of circle ${\left(y-1\right)}^2+x^2=1$ .

Graph:

The graph of circle ${\left(y-1\right)}^2+x^2=1$ by solving for $y$ .

Subtract $x^2$ from the equation of circle,

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

Take under root on both side of the equation,

  $y-1=\pm \sqrt{1-x^2}$

Add 1 on the equation,

  $\begin{array}{c}y-1+1=1\pm \sqrt{1-x^2}\\ y=1\pm \sqrt{1-x^2}\end{array}$

From the two equations graph of circle is described:

  $y=1+\sqrt{1-x^2},y=1-\sqrt{1-x^2}$

The graph of the equation $y=1+\sqrt{1-x^2}$ represents upper half of the circle because y is always greater than or equal to 1.

  $\begin{array}{c}1+\sqrt{1-x^2}\ge 1\\ \sqrt{1-x^2}\ge 0\\ 1\ge x^2\\ \pm 1\ge x\end{array}$

So, the values of $x$ is less than or equal to $\pm 1$ .

Thus, the graph appropriate in the $\left[-1\text{,}1\right]\times \left[\text{0,2}\right]$ viewing rectangle

The graph of the equation $y=1+\sqrt{1-x^2}$ can be sketched using the table,

$x$	$y$
0	2
1	1
$\frac{1}{2}$	$1+\sqrt{\frac{3}{4}}$

The graph of equation is provided below,

  

The graph of the equation $y=1-\sqrt{1-x^2}$ represents upper half of the circle because y is always less than or equal to 1.

  $\begin{array}{c}1-\sqrt{1-x^2}\le 1\\ -\sqrt{1-x^2}\le 0\\ 1-x^2\le 0\\ \pm 1\le x\end{array}$

So, the values of $x$ is greater than or equal to $\pm 1$ .

Thus, the graph appropriate in the $\left[\text{-1,1}\right]\times \left[\text{0,1}\right]$ viewing rectangle

The graph of the equation $y=1-\sqrt{1-x^2}$ can be sketched using the table,

$x$	$y$
0	0
1	1
$\frac{1}{2}$	$1-\sqrt{\frac{3}{4}}$

The graph of equation is provided below,

  

The graph of both equations in same figure is provided below,

Thus, the graph appropriate in the $\left[\text{-1,1}\right]\times \left[\text{0,2}\right]$ viewing rectangle

  

Interpretation:

The graph of an equation in a viewing screen is a viewing rectangle.

The $\text{x}$ -values to range from a minimum value of $\text{xmin}$ $=\text{a}$ to a maximum value of $\text{xmax}$ $=\text{b}$

The $\text{y}$ -values to range from minimum value of $\text{ymin}$ $=\text{c}$ to a maximum value of $\text{ymax}$ $=\text{d}$

Then, the display portion of the graph lies in the rectangle $\left[\text{a,b}\right]\times \left[\text{c,d}\right]=\left\{\left(\text{x,y}\right)|\text{a}\le \text{x}\le \text{b},\text{c}\le \text{y}\le \text{d}\right\}$

The graph of the equation $y=1+\sqrt{1-x^2}$ represents upper half of the circle appropriate in the $\left[-1\text{,}1\right]\times \left[\text{0,1}\right]$ viewing rectangle.

The graph of the equation $y=1-\sqrt{1-x^2}$ represents upper half of the circle appropriate in the $\left[-1\text{,}1\right]\times \left[\text{0,2}\right]$ viewing rectangle

The graph of the equation ${\left(y-1\right)}^2+x^2=1$ represents circle in the $\left[-1\text{,}1\right]\times \left[\text{0,2}\right]$ viewing rectangle.