# How do you find the equation of a circle with points (1, 2) and (–1, –2) as endpoints of a diameter?

Nov 25, 2016

${x}^{2} + {y}^{2} = 5$

#### Explanation:

The circle is passing through the points $\left(- 1 , - 2\right) \mathmr{and} \left(1 , 2\right)$

This is the diameter of the circle. Its midpoint is the center of the circle and half of the diameter is the radius of the circle.

Mid - point / center of the circle

$\left(x , y\right) = \frac{{x}_{1} + {x}_{2}}{2} , \frac{{y}_{1} + {y}_{2}}{2}$
$\left(x , y\right) = \frac{- 1 + 1}{2} , \frac{- 2 + 2}{2} = \left(0 , 0\right)$

Center of the circle is $\left(0 , 0\right)$

Its radius is the distance between $\left(0 , 0\right)$ and $\left(1 , 2\right)$

$r = \sqrt{\left({x}_{1} - {x}_{2}\right) + \left({y}_{1} - {y}_{2}\right)}$
r=sqrt[(0-(-1))+(0-(-2))
$r = \sqrt{1 + 4} = \sqrt{5}$

The equation of the circle having $\left(0 , 0\right)$ as the center is -

${x}^{2} + {y}^{2} = {r}^{2}$

in our case -

$r = \sqrt{5}$
Then -
$r = 5$

Hence the required equation is -

${x}^{2} + {y}^{2} = 5$