# What is the standard form of the equation of a circle with a diameter that has endpoints (-8,0) and (4,-8)?

Feb 14, 2016

${\left(x + 2\right)}^{2} + {\left(y + 4\right)}^{2} = 52$

#### Explanation:

since the coords of the endpoints of the diameter are known, the centre of the circle can be calculated using the 'mid-point formula'.The centre being at the mid-point of the diameter.

centre = $\left[\frac{1}{2} \left({x}_{1} + {x}_{2}\right) , \frac{1}{2} \left({y}_{1} + {y}_{2}\right)\right]$

let $\left({x}_{1} , {y}_{1}\right) = \left(- 8 , 0\right)$
and$\left({x}_{2} , {y}_{2}\right) = \left(4 , - 8\right)$

hence centre $= \left[\frac{1}{2} \left(- 8 + 4\right) , \frac{1}{2} \left(0 - 8\right)\right] = \left(- 2 , - 4\right)$

and radius is the distance from the centre to one of the end points. To calculate r , use the 'distance formula'.

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

let$\left({x}_{1} , {y}_{1}\right) = \left(- 2 , - 4\right)$
and$\left({x}_{2} , {y}_{2}\right) = \left(- 8 , 0\right)$

hence r $= \sqrt{{\left(- 8 + 2\right)}^{2} + {\left(0 + 4\right)}^{2}} = \sqrt{36 + 16} = \sqrt{52}$

centre = (-2 , -4 ) and $r = \sqrt{52}$

the standard form of the equation of a circle is

${\left(x - a\right)}^{2} + {\left(y - b\right)}^{2} = {r}^{2}$
where (a , b ) are the coords of centre and r , is radius.

$\Rightarrow {\left(x + 2\right)}^{2} + {\left(y + 4\right)}^{2} = 52$