Chapter 1, Problem 23RE

To determine

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

Answer to Problem 23RE

The center of the circle $\left(h,k\right)$ is (1, -2) and the radius r is 3.

  

Explanation of Solution

Given:

  $x^2+y^2-2x+4y-4=0$

Calculation:

Group the terms containing x and the terms containing y . Bring the constant to theright side of the equation.

  $\left(x^2-2x\right)+\left(y^2+4y\right)=4$

Express the terms within the parentheses as perfect squares. Any number added on the left-hand side must be added on the right-hand side also.

  $\begin{array}{l}\left(x^2-2x+1\right)+\left(y^2+4y+4\right)=4+1+4\\ \stackrel{\uparrow }{\overbrace{{\left(\frac{2}{2}\right)}^2=1}}\stackrel{\uparrow }{\overbrace{{\left(\frac{4}{2}\right)}^2=4}}\end{array}$

Factor the expression.

  ${\left(x-1\right)}^2+{\left(y+2\right)}^2=3^2$

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 equation in standard form

  ${\left(x-1\right)}^2+{\left(y+2\right)}^2=3^2$

Compare the equation with the standard form.

  $\begin{array}{l}{\left(x-h\right)}^2+{\left(y-k\right)}^2=r^2\\ \downarrow \downarrow \downarrow \\ {\left(x-1\right)}^2+{\left(y-\left(-2\right)\right)}^2=3^2\end{array}$

Therefore, the center of the circle $\left(h,k\right)$ is (1, -2) and the radius r is 3.

The center of the circle is at the point (1, -2). Since the radius is 3 units, four points on the circle can be located by plotting points that are 3 units to the right, to the left, up, and down from the center. Use these points to graph the circle.

  

Substitute $y=0$ in the standard equation to find the x -intercept.

  $\begin{array}{l}{\left(x-1\right)}^2+{\left(0+2\right)}^2=9\hfill \\ {\left(x-1\right)}^2+4=9\hfill \end{array}$

Subtract 4 from both the sides.

  $\begin{array}{l}{\left(x-1\right)}^2+4-4=9-4\\ {\left(x-\text{1}\right)}^2=5\end{array}$

Use the square root method.

  $x-1=\pm \sqrt{5}$

Add 1 to both the sides.

  $\begin{array}{l}x-1+1=\pm \sqrt{5}+1\\ x=1\pm \sqrt{5}\end{array}$

The x -intercepts are $1+\sqrt{5}$ and $1-\sqrt{5}$ .

Substitute $x=0$ in the standard equation to find the y-intercept

  $\begin{array}{l}{\left(0-1\right)}^2\text{+}{\left(y+2\right)}^2=9\hfill \\ 1+{\left(y+2\right)}^2=9\hfill \end{array}$

Subtract 1 from both the sides.

  $\begin{array}{l}1+{\left(y+2\right)}^2-1=9-1\hfill \\ {\left(y+2\right)}^2=8\hfill \end{array}$

Use the square root method.

  $y+2=\pm \sqrt{8}$

Subtract 2 from both the sides.

  $\begin{array}{l}y+2-2\text{ = }\pm \sqrt{8}-2\hfill \\ y=-2\pm \sqrt{8}\hfill \end{array}$

The y intercepts are $-2+\sqrt{8}$ and $-2-\sqrt{8}$ .

Conclusion:

Therefore, the center of the circle $\left(h,k\right)$ is (1, -2) and the radius r is 3.