How do you write an equation for a circle given endpoints of a diameter at (-5,2) and (3,6)?

Oct 24, 2016

${\left(x + 1\right)}^{2} + {\left(y - 4\right)}^{2} = 20$

Explanation:

First, we have to find the centre of the circle.

This will be halfway between the two endpoints

$\frac{3 + \left(- 5\right)}{2} = - 1$, $\frac{6 + 2}{2} = 4$

giving us the point $\left(- 1 , 4\right)$ for the centre

now we need to find the radius,
take the middle and an end point and find the distance,

((3)-(-1) , (6)-(4))

giving us $\left(4 , 2\right)$ or 4 our units horizontal and two vertical.

now if we use trigonometry to find the radius

$r = \sqrt{{4}^{2} + {2}^{2}} = \sqrt{20}$

now the equation of a circle is,

${\left(x - a\right)}^{2} + {\left(y - b\right)}^{2} = {r}^{2}$

where a and b are the coordinates of the centre.
sub in,

${\left(x - \left(- 1\right)\right)}^{2} + {\left(y - \left(4\right)\right)}^{2} = {\left(\sqrt{20}\right)}^{2}$

${\left(x + 1\right)}^{2} + {\left(y - 4\right)}^{2} = 20$

giving us,

${\left(x + 1\right)}^{2} + {\left(y - 4\right)}^{2} = 20$

for the equation of our circle.