Chapter 10.2, Problem 32E

To determine

To write: The standard form of the equation of the circle that passes through the points with the coordinates $\left(4,5\right),\left(-2,3\right),\left(-4,-3\right)$ . Then identify the center and radius.

Answer to Problem 32E

  $\begin{array}{l}{\left(x+2\right)}^2+{\left(y-3\right)}^2=50\\ r=\sqrt{50}\\ \text{Center}\left(-2,3\right)\end{array}$

Explanation of Solution

Given information: The coordinates $\left(4,5\right),\left(-2,3\right),\left(-4,-3\right)$

Calculation:

Points $\left(x,y\right):\left(4,5\right),\left(-2,3\right),\left(-4,-3\right)$

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:

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

  $\begin{array}{l}\text{Pt:}\left(4,5\right):\\ {\left(4-h\right)}^2+{\left(5-k\right)}^2=r^2\\ h^2+k^2-10h-8k+41=r^2\\ \\ \text{Pt:}\left(-2,3\right):\\ {\left(-2-h\right)}^2+{\left(3-k\right)}^2=r^2\\ h^2+k^2-6h+4k+13=r^2\\ \\ \text{Pt:}\left(-4,-3\right):\\ {\left(-4-h\right)}^2+{\left(-3-k\right)}^2=r^2\\ h^2+k^2+6h+8k+25=r^2\end{array}$

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

  $\begin{array}{l}{h}^2+k^2-10h-8k+41=h^2+k^2-6h+4k+13\\ h=-3k+7\end{array}$

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

  $\begin{array}{l}{h}^2+k^2-10h-8k+41=h^2+k^2+6k+8k+25\\ h=-k+1\end{array}$

Set these 2 results equal because they both equal r²

  $-k+1=-3k+7\to k=3\to h=-2$

You know the center $\left(h,k\right)=\left(-2,3\right)$

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

  $\begin{array}{l}{\left(-2\right)}^2+{\left(3\right)}^2-10\left(-2\right)-8\left(3\right)+41=r^2\\ r^2=50\end{array}$

Standard form:

  $\begin{array}{l}{\left(x+2\right)}^2+{\left(y-3\right)}^2=50\\ r=\sqrt{50}\\ \text{Center}\left(-2,3\right)\end{array}$