# How do you find the center-radius form of the equation of the circle having a diameter with endpoints (-5, 1) and (11, 13)?

Oct 19, 2016

${\left(x - 3\right)}^{2} + {\left(y - 7\right)}^{2} = {10}^{2}$

#### Explanation:

If $\left(- 5 , 1\right)$ and $\left(11 , 13\right)$ are diameter endpoints then the center of the circle is at the midpoint between them:
Center at $\left(\frac{- 5 + 11}{2} , \frac{1 + 13}{2}\right) = \left(\textcolor{b l u e}{3} , \textcolor{red}{7}\right)$

The radius is the distance from the center, $\left(3 , 7\right)$ to a diameter endpoint (I will use $\left(11 , 13\right)$ but either should work).

By Pythagorean Theorem
Radius $= \sqrt{{\left(11 - 3\right)}^{2} + {\left(13 - 7\right)}^{2}}$
$\textcolor{w h i t e}{\text{XXX}} = \sqrt{{8}^{2} + {6}^{2}}$
$\textcolor{w h i t e}{\text{XXX}} = \sqrt{64 + 36}$
$\textcolor{w h i t e}{\text{XXX}} = \sqrt{100}$
$\textcolor{w h i t e}{\text{XXX}} = \textcolor{g r e e n}{10}$

The center-radius form for a s circle with center at $\left(\textcolor{b l u e}{a} , \textcolor{red}{b}\right)$ and radius $\textcolor{g r e e n}{r}$ is
$\textcolor{w h i t e}{\text{XXX}} {\left(x - \textcolor{b l u e}{a}\right)}^{2} + {\left(y - \textcolor{red}{b}\right)}^{2} = {\textcolor{g r e e n}{r}}^{2}$

Using the derived values
$\textcolor{w h i t e}{\text{XXX}} {\left(x - \textcolor{b l u e}{3}\right)}^{2} + {\left(y - \textcolor{red}{7}\right)}^{2} = {\textcolor{g r e e n}{10}}^{2}$

Here's a graph of ${\left(x - 3\right)}^{2} + {\left(y - 7\right)}^{2} = {10}^{2}$ for verification purposes: