# How do you write the equation of the circle with diameter that has endpoints at (–1, –6) and (–3, –2).?

Jun 30, 2016

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

#### Explanation:

The center of the circle is the middle point of the diameter.

Then its coordinates are obtained by using

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

in this case:

((-1-3)/2;(-6-2)/2)

(-2;-4)

The radius is the half of the diameter, that is obtained by calculating the distance

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

in this case:

$\sqrt{{\left(- 3 + 1\right)}^{2} + {\left(- 2 + 6\right)}^{2}}$

$\sqrt{4 + 16} = \sqrt{20}$

To obtain teh equation of the circle you can substitute in the general equation ${x}_{0} = - 2 , {y}_{0} = - 4 , r = \sqrt{20}$:

(the general equation is ${\left(x - {x}_{0}\right)}^{2} + {\left(y - {y}_{0}\right)}^{2} = {r}^{2}$)

${\left(x + 2\right)}^{2} + {\left(y + 4\right)}^{2} = 20$

${x}^{2} + 4 + 4 x + {y}^{2} + 16 + 8 y - 20 = 0$

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