Chapter 10.2, Problem 10CFU

To determine

To write: The standard form of the equation of the circle that passes through points with the coordinates $\left(0,0\right),\left(4,0\right),\left(0,4\right)$ . Then identify the center and the radius.

Answer to Problem 10CFU

  $\begin{array}{l}{\left(x+2\right)}^2+{\left(y-2\right)}^2=8\\ \text{Center:}\left(-2,2\right)\\ r=2\sqrt{2}\end{array}$

Explanation of Solution

Given information: The coordinates $\left(0,0\right),\left(4,0\right),\left(0,4\right)$

Calculation:

Points $\left(x,y\right):\left(0,0\right),\left(4,0\right),\left(0,4\right)$

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:

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

  $\begin{array}{l}\text{Pt:}\left(0,0\right):\\ {\left(0-h\right)}^2+{\left(0-k\right)}^2=r^2\\ h^2+k^2=r^2\\ \\ \text{Pt:}\left(4,0\right):\\ {\left(-4-h\right)}^2+{\left(0-k\right)}^2=r^2\\ h^2+k^2+8h+16=r^2\\ \\ \text{Pt:}\left(0,4\right):\\ {\left(0-h\right)}^2+{\left(4-k\right)}^2=r^2\\ h^2+k^2-8k+16=r^2\end{array}$

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

  $\begin{array}{l}{h}^2+k^2=h^2+k^2+8h+16\\ 8h+16=0\to h=-2\end{array}$

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

  $\begin{array}{l}{h}^2+k^2=h^2+k^2-8k+16\\ -8k+16=0\to k=2\end{array}$

Use these values to find $r^2$ .

  $\begin{array}{l}{2}^2+{\left(-2\right)}^2=r^2\to r^2=8\\ r=2\sqrt{2}\end{array}$

Final equation:

  $\begin{array}{l}{\left(x+2\right)}^2+{\left(y-2\right)}^2=8\\ \text{Center:}\left(-2,2\right)\\ r=2\sqrt{2}\end{array}$