Chapter 1, Problem 24RE

To determine

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

Answer to Problem 24RE

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

  

Explanation of Solution

Given:

  $x^2+y^2+4x-4y-1=0$

Calculation:

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

  $\left(x^2+4x\right)+\left(y^2-4y\right)=1$

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.

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

Factor the expression.

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

The standard equation of a circle with center $\left(h,k\right)$ and radius r is

  ${\left(x-h\right)}^2+{\left(y-k\right)}^2=r^2$ .

Compare the equation ${\left(x+2\right)}^2+{\left(y-2\right)}^2=9$ with the standard form.

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

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

The center of the circle is at the point $\left(-2,1\right)$ . Since the radius is 3, move 3 units to the left from the center and plot the corresponding point. Similarly, plot points 3 units to the right, up, and down from the center and draw a smooth curve through these points.

  

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

  $\begin{array}{l}{\left(x+2\right)}^2+{\left(y-1\right)}^2=3^2\\ {\left(x+2\right)}^2+1=9\end{array}$

Subtract 1 from both the sides.

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

Use the square root method.

  $\begin{array}{l}\sqrt{{\left(x+2\right)}^2}=\sqrt{8}\\ x+2=\pm \sqrt{8}\end{array}$

Solve for x,

  $x=-2\pm \sqrt{8}$

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

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

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

Subtract 4 from both the sides.

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

Use the square root method.

  $\begin{array}{l}\sqrt{{\left(y-1\right)}^2}=\sqrt{5}\\ y-1=\pm \sqrt{5}\\ y=\pm \sqrt{5}-1\end{array}$

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

Conclusion:

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