Chapter 1, Problem 25RE

To determine

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

Answer to Problem 25RE

The center of the circle $\left(h,k\right)$ is $\left(1,-2\right)$ and the radius r is $\sqrt{5}$ .

  

Explanation of Solution

Given:

  $3x^2+3y^2-6x+12y=0$

Calculation:

Factor out the common factor 3.

  $3\left(x^2+y^2-2x+4y\right)=0$

By zero product property $x^2+y^2-2x+4y=0$

Group the terms containing x and the terms containing y .

  $\left(x^2-2x\right)+\left(y^2+4y\right)=0$

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.

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

Factor the expression.

  ${\left(x-1\right)}^2+{\left(y+2\right)}^2=5$

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

  ${\left(x-1\right)}^2+{\left(y-\left(-2\right)\right)}^2={\left(\sqrt{5}\right)}^2$

Compare the equation with the standard form.

  $\begin{array}{l}{\left(x-h\right)}^2+{\left(y-k\right)}^2=r^2\\ \downarrow \downarrow \downarrow \\ {\left(x-1\right)}^2+{\left(y-\left(-2\right)\right)}^2={\left(\sqrt{5}\right)}^2\end{array}$

Therefore, the center of the circle $\left(h,k\right)$ is $\left(1,-2\right)$ and the radius ris $\sqrt{5}$ .

The center of the circle is at the point $\left(1,-2\right)$ . Since the radius is $\sqrt{5}$ units, four points on the circle can be located by plotting points that are 1 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.

  $\begin{array}{l}{\left(x-1\right)}^2+{\left(0+2\right)}^2=5\\ {\left(x-1\right)}^2+4=5\end{array}$

Subtract 4 from both the sides.

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

Use the square root method.

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

Add 1 to both the sides,

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

The x -intercepts are 2 and 0.

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

  $\begin{array}{l}{\left(0-1\right)}^2+{\left(y+2\right)}^2=5\\ 1+{\left(y+2\right)}^2=5\end{array}$

Subtract 1 from both the sides.

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

Use the square root method.

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

Subtract 2 from both sides,

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

The y - intercepts are 0 and-4.

Conclusion:

Therefore, the center of the circle $\left(h,k\right)$ is $\left(1,-2\right)$ and the radius r is $\sqrt{5}$ .