# How do you find the equation of the circle with a diameter that has endpoints at (-7, -2) and (-15, 6)?

Dec 28, 2015

${\left(x + 11\right)}^{2} + {\left(y - 2\right)}^{2} = 128$

#### Explanation:

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

Where $C : \left(h , k\right)$

To find the center, use the midpoint formula

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

$\implies C : \left(\frac{- 7 + - 15}{2} , \frac{- 2 + 6}{2}\right)$

$\implies C : \left(- 11 , 2\right)$

To find the radius, use the distance formula

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

$\implies r = \sqrt{{\left(- 7 - - 15\right)}^{2} + {\left(- 2 - 6\right)}^{2}}$

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

=> r = sqrt(128

$\implies r = 8 \sqrt{2}$

Therefore, the equation of the circle is

${\left(x + 11\right)}^{2} + {\left(y - 2\right)}^{2} = {\left(8 \sqrt{2}\right)}^{2}$

$\implies {\left(x + 11\right)}^{2} + {\left(y - 2\right)}^{2} = 128$