# How do you write the equation for a circle where the points (2, 6) and (8, 10) lie along a diameter?

May 21, 2016

${\left(x - 5\right)}^{2} + {\left(y - 8\right)}^{2} = 13$

#### Explanation:

If the points $\left(\textcolor{b r o w n}{2} , \textcolor{b r o w n}{6}\right)$ and $\left(\textcolor{t e a l}{8} , \textcolor{t e a l}{10}\right)$ are end points of a diameter of a circle,
the center of the circle is at $\left(\frac{\textcolor{b r o w n}{2} + \textcolor{t e a l}{8}}{2} , \frac{\textcolor{b r o w n}{6} + \textcolor{t e a l}{10}}{2}\right) = \left(\textcolor{red}{5} , \textcolor{b l u e}{8}\right)$

and the radius of the circle is $\textcolor{g r e e n}{r} = \sqrt{{\left(\textcolor{red}{5} - \textcolor{b r o w n}{2}\right)}^{2} + {\left(\textcolor{b l u e}{8} - \textcolor{b r o w n}{6}\right)}^{2}} = \textcolor{g r e e n}{\sqrt{13}}$

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

So, in this case, the equation of the required circle is
$\textcolor{w h i t e}{\text{XXX}} {\left(x - \textcolor{red}{5}\right)}^{2} + {\left(y - \textcolor{b l u e}{8}\right)}^{2} = \textcolor{g r e e n}{13}$