Chapter 10.2, Problem 44E

To determine

To write: The equation of circle of the given details.

Answer to Problem 44E

  ${\left(x-\frac{13}{2}\right)}^2+{\left(y-\frac{9}{2}\right)}^2=\frac{65}{2}$

Explanation of Solution

Given information: Geometry: A circle that circumscribes the triangle whose sides are the graphs of $4x-7y=27,x-5y+3=0\text{ and }2x+3y-7=0$ .

Calculation:

If you graph the lines, you need to find the points of intersection. This is the first step.

You can solve them with the calculator, substitution or elimination. This gives you 3 points that the circle passes through.

From here this problem is similar to problems solved earlier.

  

Points $\left(x,y\right):\left(2,1\right),\left(5,-1\right),\left(12,3\right)$

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:

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

  $\begin{array}{l}\text{Pt:}\left(2,1\right):\\ {\left(2-h\right)}^2+{\left(1-k\right)}^2=r^2\\ h^2+k^2-4h-2k+5=r^2\\ \\ \text{Pt:}\left(5,-1\right):\\ {\left(5-h\right)}^2+{\left(-1-k\right)}^2=r^2\\ h^2+k^2-10h+2k+26=r^2\\ \\ \text{Pt:}\left(12,3\right):\\ {\left(12-h\right)}^2+{\left(3-k\right)}^2=r^2\\ h^2+k^2-24h-6k+153=r^2\end{array}$

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

  $\begin{array}{l}{h}^2+k^2-4h-2k+5=h^2+k^2-10h+2k+26\\ 6h=4k+21\to h=\frac{4k+21}{6}\end{array}$

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

  $\begin{array}{l}{h}^2+k^2-4h-2k+5=h^2+k^2-24h-6k+153\\ 20h=-4k+148=-\frac{4k-148}{20}\end{array}$

Set the 2 equations for h equal to find k .

  $\begin{array}{l}-\frac{4k-148}{20}=\frac{4k+21}{6}\to k=\frac{9}{2}\\ \text{This means:}h=\frac{4\left(\frac{9}{2}\right)+21}{6}=\frac{13}{2}\end{array}$

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

  $\begin{array}{l}{\left(\frac{13}{2}\right)}^2+{\left(\frac{9}{2}\right)}^2-4\left(\frac{13}{2}\right)-2\left(\frac{9}{2}\right)+5=r^2\\ r^2=\frac{65}{2}\end{array}$

Final equation is:

  ${\left(x-\frac{13}{2}\right)}^2+{\left(y-\frac{9}{2}\right)}^2=\frac{65}{2}$