# How do you find the center-radius form of the equation of the circle having a diameter with endpoints (0, -8) and (-6,0)?

Jan 9, 2017

Use $h = \frac{{x}_{0} + {x}_{1}}{2} \mathmr{and} k = \frac{{y}_{0} + {y}_{1}}{2}$to find the center. Substitute the center and one of the points into the general form, to find the radius.

#### Explanation:

Find the center:

$h = \frac{0 + - 6}{2} = - 3$

$k = \frac{- 8 + 0}{2} = - 4$

Substitute the center and a point into the equation for a circle:

${\left(0 - - 3\right)}^{2} + {\left(- 8 - - 4\right)}^{2} = {r}^{2}$

Solve for r:

$9 + 16 = {r}^{2}$

${r}^{2} = 25$

$r = 5$

The equation of the circle is:

${\left(x - - 3\right)}^{2} + {\left(y - - 4\right)}^{2} = {5}^{2}$