# Find the equation to the circle such that the points A(-3,5) and B(4,-2) form the ends of a diameter?

## this is all the given information.

Mar 25, 2018

${\left(x - \frac{1}{2}\right)}^{2} + {\left(y - \frac{3}{2}\right)}^{2} = \frac{49}{2}$

#### Explanation:

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

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

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

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

$\text{the coordinates of the midpoint are the average of the}$
$\text{the coordinates of the endpoints}$

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

$\text{to calculate the radius use the "color(blue)"distance formula}$

•color(white)(x)r=sqrt((x_2-x_1)^2+(y_2-y_1)^2)

$\text{let "(x_1,y_1)=(1/2,3/2)" and } \left({x}_{2} , {y}_{2}\right) = \left(- 3 , 5\right)$

rArrr=sqrt((-3-1/2)^2+(5-3/2)^2

$\textcolor{w h i t e}{r} = \sqrt{\frac{49}{4} + \frac{49}{4}} = \sqrt{\frac{98}{4}}$

$\Rightarrow {\left(x - \frac{1}{2}\right)}^{2} + {\left(y - \frac{3}{2}\right)}^{2} = {\left(\sqrt{\frac{98}{4}}\right)}^{2}$

$\Rightarrow {\left(x - \frac{1}{2}\right)}^{2} + {\left(y - \frac{3}{2}\right)}^{2} = \frac{49}{2} \leftarrow \textcolor{red}{\text{equation of circle}}$