# How do you write the standard form of the equation of the circle with the given the diameter with endpoints (-10, -6) and (-2, -4)?

Nov 6, 2016

${\left(x + 6\right)}^{2} + {\left(y + 5\right)}^{2} = 17$

#### Explanation:

Diameter with endpoints $\left(- 10 , - 6\right)$ and $\left(- 2 , - 4\right)$

To find the length of the diameter, use the distance formula

$d = \sqrt{{\left({x}_{1} - {x}_{2}\right)}^{2} + {\left({y}_{1} - {y}_{2}\right)}^{2}}$
where $d$ is the distance between two points $\left({x}_{1} , {y}_{1}\right)$ and $\left({x}_{2} , {y}_{2}\right)$

$d = \sqrt{{\left(- 10 - - 2\right)}^{2} + {\left(- 6 - - 4\right)}^{2}}$

$d = \sqrt{{\left(- 8\right)}^{2} + {\left(- 2\right)}^{2}}$

$d = \sqrt{64 + 4}$

$d = \sqrt{68} = 2 \sqrt{17}$

The length of the radius $r$ equals the diameter divided by 2.

$r = \frac{2 \sqrt{17}}{2} = \sqrt{17}$

The center of the circle is the midpoint of the diameter.

Remember, the midpoint of a segment is just the average of the x coordinates and the average of the y coordinates. This gives the midpoint formula

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

midpoint $= \left(\frac{- 10 + - 2}{2} , \frac{- 6 + - 4}{2}\right) = \left(- 6 , - 5\right)$

The center is $\left(- 6 , - 5\right)$ and the radius is $\sqrt{17}$

The equation of a circle is ${\left(x - h\right)}^{2} + {\left(y - k\right)}^{2} = {r}^{2}$,
where $\left(h , k\right)$ is the center and $r$ is the radius.

Therefore, the equation is

${\left(x - - 6\right)}^{2} + {\left(y - - 5\right)}^{2} = {\left(\sqrt{17}\right)}^{2}$

${\left(x + 6\right)}^{2} + {\left(y + 5\right)}^{2} = 17$