# How do you find the equation of the circle with a diameter that has endpoints (-8,0) and (4,-8)?

Sep 4, 2016

Equation of circle is
${x}^{2} + {y}^{2} + 4 x + 8 y - 32 = 0$

#### Explanation:

As the end points of diameter are $\left(- 8 , 0\right)$ and $\left(4 , - 8\right)$, their midpoint is center given by $\left(\frac{- 8 + 4}{2} , \frac{0 - 8}{2}\right)$ i.e. $\left(- 2 , - 4\right)$.

The distance between $\left(- 8 , 0\right)$ and $\left(4 , - 8\right)$ is given by

sqrt((4-(-8))^2+(-8-0)^2))

= $\sqrt{144 + 64}$

= sqrt208=sqrt(2x2x2×2×13)

= $4 \sqrt{13}$

Hence radius is $2 \sqrt{13}$

As center is $\left(- 2 , - 4\right)$ and radius is $2 \sqrt{13}$, the equation of circle is

${\left(x - \left(- 2\right)\right)}^{2} + {\left(y - \left(- 4\right)\right)}^{2} = {\left(2 \sqrt{13}\right)}^{2}$ or

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

${x}^{2} + 4 x + 4 + {y}^{2} + 8 y + 16 = 52$ or

${x}^{2} + {y}^{2} + 4 x + 8 y - 32 = 0$