# How do you write the equation for a circle with the endpoints of a diameter at (4, -3) and (-2, 5)?

Apr 25, 2018

${\left(x - 1\right)}^{2} + {\left(y - 1\right)}^{2} = 25$

#### Explanation:

$\text{the equation of a circle in standard form is.}$

$\textcolor{red}{\overline{\underline{| \textcolor{w h i t e}{\frac{2}{2}} \textcolor{b l a c k}{{\left(x - a\right)}^{2} + {\left(y - b\right)}^{2} = {r}^{2}} \textcolor{w h i t e}{\frac{2}{2}} |}}}$

$\text{where "(a,b)"are the coordinates of the centre and r}$
$\text{is the radius}$

$\text{we require to find the centre and radius}$

$\text{given the endpoints of the diameter then the centre}$
$\text{is at the midpoint and the radius is the distance from}$
$\text{the centre to either of the endpoints}$

$\text{midpoint } = \left[\frac{1}{2} \left(4 - 2\right) , \frac{1}{2} \left(5 - 3\right)\right]$

$\Rightarrow \text{centre } = \left(1 , 1\right)$

$\text{using "(1,1)" and } \left(4 , - 3\right)$

$r = \sqrt{{\left(4 - 1\right)}^{2} + {\left(- 3 - 1\right)}^{2}} = \sqrt{9 + 16} = 5$

${\left(x - 1\right)}^{2} + {\left(y - 1\right)}^{2} = 25 \leftarrow \textcolor{red}{\text{equation of circle}}$