# How do you write the standard form of the equation of the circle whose diameter has endpoints of (-2, 4) and (4, 12)?

${\left(x - 1\right)}^{2} + {\left(y - 8\right)}^{2} = 25$

#### Explanation:

The given data are the endpoints ${E}_{1} \left({x}_{1} , {y}_{1}\right) = \left(- 2 , 4\right)$ and ${E}_{2} \left({x}_{2} , {y}_{2}\right) = \left(4 , 12\right)$ of the diameter $D$ of the circle

Solve for the center $\left(h , k\right)$

$h = \frac{{x}_{1} + {x}_{2}}{2} = \frac{- 2 + 4}{2} = 1$
$k = \frac{{y}_{1} + {y}_{2}}{2} = \frac{4 + 12}{2} = 8$

Center $\left(h , k\right) = \left(1 , 8\right)$

Solve now for the radius $r$

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

$r = \frac{D}{2} = \frac{\sqrt{{\left(- 2 - 4\right)}^{2} + {\left(4 - 12\right)}^{2}}}{2}$

$r = \frac{D}{2} = \frac{\sqrt{36 + 64}}{2}$

$r = \frac{D}{2} = \frac{\sqrt{100}}{2}$

$r = \frac{D}{2} = \frac{10}{2}$

$r = 5$

The standard form of the equation of the circle:

${\left(x - h\right)}^{2} + {\left(y - k\right)}^{2} = {r}^{2}$
${\left(x - 1\right)}^{2} + {\left(y - 8\right)}^{2} = {5}^{2}$
${\left(x - 1\right)}^{2} + {\left(y - 8\right)}^{2} = 25$