Chapter 13.2, Problem 25PSC

To determine

To find: The center of the circle containing $D=\left(-3,5\right)$ , $E=\left(3,3\right)$ and $F=\left(11,19\right)$ .

Answer to Problem 25PSC

The center of the circle containing $D=\left(-3,5\right)$ , $E=\left(3,3\right)$ and $F=\left(11,19\right)$ is $\left(3,13\right)$ .

Explanation of Solution

Given:

The points are $D=\left(-3,5\right)$ , $E=\left(3,3\right)$ and $F=\left(11,19\right)$ .

Calculation:

The perpendicular bisector of chords of a point passes through the center.

Find the mid-point of the line $\overleftrightarrow{\text{DE}}$ .

  $\begin{array}{c}\text{Midpoint}=\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)\\ =\left(\frac{3-3}{2},\frac{5+3}{2}\right)\\ =\left(0,4\right)\end{array}$

Find the mid-point of the line $\overleftrightarrow{\text{EF}}$ .

  $\begin{array}{c}\text{Midpoint}=\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)\\ =\left(\frac{11+3}{2},\frac{19+3}{2}\right)\\ =\left(7,11\right)\end{array}$

Find the slope of line $\overleftrightarrow{\text{DE}}$ .

  $\begin{array}{c}m=\frac{y_2-y_1}{x_2-x_1}\\ m=\frac{3-5}{3+3}\\ =-\frac{1}{3}\end{array}$

So, the slope of perpendicular bisector will be $3$ .

Find the slope of line $\overleftrightarrow{\text{EF}}$ .

  $\begin{array}{c}m=\frac{y_2-y_1}{x_2-x_1}\\ m=\frac{19-3}{11-3}\\ =\frac{16}{8}\\ =2\end{array}$

So, the slope of perpendicular bisector will be $-\frac{1}{2}$ .

So, the slope of perpendicular bisector will be $3$ .

Find the equation of the perpendicular bisector of $\overleftrightarrow{\text{DE}}$ .

  $\begin{array}{c}y-y_1=m\left(x-x_1\right)\\ y-4=3\left(x-0\right)\\ y-4=3x\\ y=3x+4\end{array}$

Find the equation of the perpendicular bisector of $\overleftrightarrow{\text{EF}}$ .

  $\begin{array}{c}y-y_1=m\left(x-x_1\right)\\ y-11=-\frac{1}{2}\left(x-7\right)\\ y-11=-\frac{1}{2}x+\frac{7}{2}\\ y=-\frac{1}{2}x+\frac{29}{2}\end{array}$

The intersection of the equations $y=3x+4$ , and $y=-\frac{1}{2}x+\frac{29}{2}$ is the center.

  $\begin{array}{c}3x+4=-\frac{1}{2}x+\frac{29}{2}\\ 6x+8=-x+29\\ 7x=21\\ x=3\end{array}$

Substitute $3$ for $x$ in the equation $y=3x+4$ .

  $\begin{array}{c}y=3\left(3\right)+4\\ =9+4\\ =13\end{array}$

Therefore, the center of the circle containing $D=\left(-3,5\right)$ , $E=\left(3,3\right)$ and $F=\left(11,19\right)$ is $\left(3,13\right)$ .