# How do you write the equation of the circle with end points of the circle are (-3, 5) (9, -7)?

Jul 22, 2017

${\left(x - 3\right)}^{2} + {\left(y + 1\right)}^{2} = 72$

#### Explanation:

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

$\textcolor{red}{\overline{\underline{| \textcolor{w h i t e}{\frac{2}{2}} \textcolor{b l a c k}{{\left(x - a\right)}^{2} + {\left(y - b\right)}^{2} = {r}^{2}} \textcolor{w h i t e}{\frac{2}{2}} |}}}$
where ( a , b ) are the coordinates of the centre and r the radius.

$\text{the centre of the circle is at the midpoint of the 2 given}$
$\text{points and the radius is the distance from the centre to}$
$\text{either of the 2 points}$

$\text{midpoint } = \left[\frac{1}{2} \left(- 3 + 9\right) , \frac{1}{2} \left(5 - 7\right)\right] = \left(3 , - 1\right)$

$\text{calculate distance between "(3,-1)" and } \left(- 3 , 5\right)$

$\text{using the "color(blue)"distance formula}$

$r = \sqrt{{\left(- 3 - 3\right)}^{2} + {\left(5 + 1\right)}^{2}} = \sqrt{72}$

$\text{using " (a,b)=(3,-1)" and } r = \sqrt{72}$

$\Rightarrow {\left(x - 3\right)}^{2} + {\left(y + 1\right)}^{2} = 72 \text{ is the equation}$