Chapter 10.2, Problem 29E

To determine

To write: The standard form of the equation of the circle that passes through the points with the coordinates $\left(7,-1\right),\left(11,-5\right),\left(3,-5\right)$ . Then identify the center and radius.

Answer to Problem 29E

  $\begin{array}{l}{\left(x-7\right)}^2+{\left(y+5\right)}^2=16\\ r=4\\ \text{Center}\left(7,-5\right)\end{array}$

Explanation of Solution

Given information: The coordinates $\left(7,-1\right),\left(11,-5\right),\left(3,-5\right)$

Calculation:

Points $\left(x,y\right):\left(7,-1\right),\left(11,-5\right),\left(3,-5\right)$

Plug in the 3 points to create 3 equations which we can use to solve the system to find all of the missing values.

Remember the form:

  $\begin{array}{l}{\left(x-h\right)}^2+{\left(y-k\right)}^2=r^2\\ \text{Vertex at:}\left(h,k\right)\end{array}$

  $\begin{array}{l}\text{Pt:}\left(7,-1\right):\\ {\left(7-h\right)}^2+{\left(-1-k\right)}^2=r^2\\ h^2+k^2-14h+2k+50=r^2\\ \\ \text{Pt:}\left(11,-5\right):\\ {\left(11-h\right)}^2+{\left(-5-k\right)}^2=r^2\\ h^2+k^2-22h+10k+146=r^2\\ \\ \text{Pt:}\left(3,-5\right):\\ {\left(3-h\right)}^2+{\left(-5-k\right)}^2=r^2\\ h^2+k^2-6h+10k+34=r^2\end{array}$

Set 1^st 2 equations equal to each other: because they both equal $r^2$ .

  $\begin{array}{l}{h}^2+k^2-14h+2k+50=h^2+k^2-22h+10k+146\\ 8h=8k+96\to h=k+12\end{array}$

Set the 1^st and 3^rd equations equal: because they also both equal $r^2$ .

  $\begin{array}{l}{h}^2+k^2-14h+2k+50=h^2+k^2-6h+10k+34\\ 8h=-8k+16\to h=2-k\end{array}$

Set these 2 results equal because they both = h

  $k+12=2-k\to k=-5$

  $\text{If }k=-5\text{ then }h=7$

You know the center $\left(h,k\right)=\left(7,-5\right)$

Plug these values for h and k into any of the equations to find r .

  $\begin{array}{l}{7}^2+{\left(-5\right)}^2-14\left(7\right)+2\left(-5\right)+50=r^2\\ r^2=16\end{array}$

Standard form:

  $\begin{array}{l}{\left(x-7\right)}^2+{\left(y+5\right)}^2=16\\ r=4\\ \text{Center}\left(7,-5\right)\end{array}$