# What is the standard form of the equation of a circle with endpoints of the diameter at (0,10) and (-10,-2)?

Mar 24, 2016

${\left(x + 5\right)}^{2} + {\left(y - 4\right)}^{2} = 61$

#### Explanation:

The equation of a circle in standard form is

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

where
$h$: $x$-coordinate of the center
$k$: $y$-coordinate of the center
$r$: radius of the circle

To get the center, get the midpoint of the endpoints of the diameter

#h = (x_1 + x_2)/2

$\implies h = \frac{0 + - 10}{2}$
$\implies h = - 5$

$k = \frac{{y}_{1} + {y}_{2}}{2}$

$\implies k = \frac{10 + - 2}{2}$
$\implies k = 4$

$c : \left(- 5 , 4\right)$

To get the radius, get the distance between the center and either endpoint of the diameter

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

$r = \sqrt{{\left(0 - - 5\right)}^{2} + {\left(10 - 4\right)}^{2}}$

$r = \sqrt{{5}^{2} + {6}^{2}}$

$r = \sqrt{61}$

Hence, the equation of the circle is

${\left(x - - 5\right)}^{2} + {\left(y - 4\right)}^{2} = {\left(\sqrt{61}\right)}^{2}$

$\implies {\left(x + 5\right)}^{2} + {\left(y - 4\right)}^{2} = 61$