# How do you write an equation for a circle with (-10 , 0) to (-16 , -10) as a diameter?

Oct 26, 2016

The equation is ${\left(x + 13\right)}^{2} + {\left(y + 5\right)}^{2} = 34$

#### Explanation:

Start by finding the length of the diameter using the distance formula.

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

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

$d = \sqrt{136}$

$d = 2 \sqrt{34}$

We will now find the radius using the formula $d = 2 r$

$2 \sqrt{34} = 2 r$

$\sqrt{34} = r$

Now, let's find the centre of the circle. We can do this using the midpoint formula.

Let $c$ be the centre:

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

$c = \left(\frac{- 10 + \left(- 16\right)}{2} , \frac{- 10 + 0}{2}\right)$

$c = \left(- \frac{26}{2} , - \frac{10}{2}\right)$

$c = \left(- 13 , - 5\right)$

We can now write the equation of the circle. The form ${\left(x - a\right)}^{2} + {\left(x - b\right)}^{2} = {r}^{2}$ represents the equation of the circle, where the point $\left(a , b\right)$ is the centre and $r$ is the radius.

Substituting, we get:

${\left(x - \left(- 13\right)\right)}^{2} + {\left(y - \left(- 5\right)\right)}^{2} = {\left(\sqrt{34}\right)}^{2}$

${\left(x + 13\right)}^{2} + {\left(y + 5\right)}^{2} = 34$

Hopefully this helps!