# How do you write an equation of a circle whose diameter of the circle are (-2,6) and (-8,10)?

Dec 3, 2016

${\left(x - - 5\right)}^{2} + {\left(y - 8\right)}^{2} = {\left(\sqrt{13}\right)}^{2}$

#### Explanation:

The general equation of a circle is:

${\left(x - h\right)}^{2} + {\left(y - k\right)}^{2} = {r}^{2} \text{ [1]}$

where $\left(x , y\right)$ is any point on the circle, $\left(h , k\right)$ is the center, and r is the radius.

To go from $\left(- 2 , 6\right) \text{ to } \left(- 8 , 10\right)$ one must move to the left 6 and up 4, therefore, the center must be to the left 3 and up 2; the point $\left(- 5 , 8\right)$. Substitute the center into equation [1]:

${\left(x - - 5\right)}^{2} + {\left(y - 8\right)}^{2} = {r}^{2} \text{ [2]}$

Substitute the point $\left(- 2 , 6\right)$ into equation [2]:

${\left(- 2 - - 5\right)}^{2} + {\left(6 - 8\right)}^{2} = {r}^{2}$

Solve for r:

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

$13 = {r}^{2}$

$r = \sqrt{13}$

Substitute $r = \sqrt{13}$ into equation [2]:

${\left(x - - 5\right)}^{2} + {\left(y - 8\right)}^{2} = {\left(\sqrt{13}\right)}^{2}$

This is the equation of the circle.