Chapter 1, Problem 21RE

To determine

To find: the center and radius of each circle and graph the circle.

Answer to Problem 21RE

The center of the circle $\left(h,k\right)$ is $\left(0,1\right)$ and the radius r is 2.

  

Explanation of Solution

Given:

  $x^2+{\left(y-1\right)}^2=4$

Calculation:

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

Express the given equation in the standard form.

  ${\left(x-0\right)}^2+{\left(y-1\right)}^2=2^2$

Compare the equation with the standard form.

  $\begin{array}{l}{\left(x-h\right)}^2+{\left(y-k\right)}^2=r^2\\ \downarrow \downarrow \downarrow \\ {\left(x-0\right)}^2+{\left(y-1\right)}^2=2^2\end{array}$

Therefore, the center of the circle $\left(h,k\right)$ is $\left(0,1\right)$ and the radius r is 2.

The center of the circle is at the point $\left(0,1\right)$ . Since the radius is 2 units, four points on the circle can be located by plotting points that are 2 units to the right to the left, up, and down from the center. Use these points to graph the circle.

  

Substitute 0for y in ${\left(x-0\right)}^2+{\left(y-1\right)}^2=2^2$ to find the x -intercept.

  $\begin{array}{l}{x}^2+{\left(0-1\right)}^2=4\\ x^2+1=4\end{array}$

Subtract 1 from both the sides.

  $\begin{array}{l}{x}^2+1-1=4-1\\ x^2=3\end{array}$

Use the square root method.

  $x=\pm \sqrt{3}$

The x -intercepts are $\sqrt{3}$ and $-\sqrt{3}$ .

Substitute 0 for x in ${\left(x-0\right)}^2+{\left(y-1\right)}^2=2^2$ to find the y -intercept.

  $\begin{array}{l}{0}^2+{\left(y-1\right)}^2=4\\ {\left(y-1\right)}^2=4\end{array}$

Use the square root method.

  $\begin{array}{l}y-1=\sqrt{4}\\ y-1=\pm 2\end{array}$

Add 1 to both the sides.

  $\begin{array}{l}y-1+1=\pm 2+1\\ y=\pm 2+1\end{array}$

The y-intercepts are -1 and 3.

Therefore the intercepts are $\left(\pm \sqrt{3},0\right),\left(0,-1\right)and\left(0,3\right)$ .

Conclusion:

Therefore the center of the circle $\left(h,k\right)$ is $\left(0,1\right)$ and the radius r is 2.