Q 22. - Precalculus Enhanced with Graphing Utilities

Question

In Problems, (a) find the center $\left(h,k\right)$ and radius r of each circle; (b) graph each circle; (c) find the intercepts, if any.

$x^2+(y-1{)}^2=1$

Step-by-Step Solution

Answer

Part (a) Center is $\left(0,1\right)$ and radius is $1$ units

Part (b) Graph is as follows:

(c) $x-$intercept is $\left(0,0\right)$ and $y-$intercepts are $\left(0,0\right),\left(0,2\right)$

Step 1. Given information

It is given that $x^2+(y-1{)}^2=1$.

Step 2. Find the center and the radius.

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

$\begin{array}{rcl}{x}^2+(y-1{)}^2& =& 1\\ (x-0{)}^2+(y-1{)}^2& =& 1^2\end{array}$

The above equation is standard form of the circle with radius $1$ and center $\left(0,1\right)$

Step 3. Graph of the circle

Graph is as follows:

Step 4. Find the intercepts.

Consider the equation $x^2+(y-1{)}^2=1$

To find the $x$-intercepts, substitute $y=0$ and solve for $x$.

$\begin{array}{rcl}{x}^2+(0-1{)}^2& =& 1\\ x^2+1& =& 1\end{array}$

Subtract 1 from both sides

$x^2=0$

Take square root of both sides

$\begin{array}{rcl}x& =& \pm \sqrt{0}\\ x& =& 0\end{array}$

Therefore, the $x$-intercept is $(0,0)$

To find the $y$-intercepts, substitute $x=0$ and solve for $y$

$\begin{array}{rcl}{0}^2+(y-1{)}^2& =& 1\\ (y-1{)}^2& =& 1\end{array}$

Take square root of both sides.

$y-1=\pm \sqrt{1}$

Add $1$ on both sides.

$$\begin{gathered} y=1\pm 1 \\ y=0,2 \end{gathered}$$

Therefore, the $y$-intercepts are $(0,0)$ and $(0,2)$.