# How do you find the general form of the equation of this circle with end points of a diameter at (1,4) and (-3,2)?

Dec 14, 2015

${\left(x + 1\right)}^{2} + \left(y - 3\right) = 5$

#### Explanation:

If the end points of a diameter of a circle are at $\left(1 , 4\right)$ and $\left(- 3 , 2\right)$
then the center of the circle is at
$\textcolor{w h i t e}{\text{XXX}} \left(\frac{1 + \left(- 3\right)}{2} , \frac{4 + 2}{2}\right) = \left(- 1 , 3\right)$
and the radius of the circle is
$\textcolor{w h i t e}{\text{XXX}} \sqrt{{\left(1 - \left(- 1\right)\right)}^{2} + {\left(4 - 3\right)}^{2}} = \sqrt{5}$

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

The required circle has an equation of
$\textcolor{w h i t e}{\text{XXX}} {\left(x + 1\right)}^{2} + {\left(y - 3\right)}^{2} = 5$
graph{(x+1)^2+(y-3)^2=5 [-5.47, 4.396, 0.33, 5.264]}