# How do you write an equation for a circle with endpoints of the diameter at (2, -5) and (-4, 3)?

Aug 11, 2016

${x}^{2} + {y}^{2} + 2 x + 2 y - 23 = 0$

#### Explanation:

To obtain the radius you first can find the lenght of the diameter by calculating the distance between its endpoints:

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

$d = \sqrt{{\left(3 + 5\right)}^{2} + {\left(- 4 - 2\right)}^{2}}$

$= \sqrt{64 + 36} = 10$

the radius is then the half of the diameter:

$r = 5$

The center of the circle is the middlepoint of the diameter:

M=((x_1+x_2)/2;(y_1+y_2)/2)

M=((2-4)/2;(-5+3)/2)=(-1;-1)

The equation of the circle is obtained by substituting the center and the radius in the following:

${\left(x - {x}_{0}\right)}^{2} + {\left(y - {y}_{0}\right)}^{2} = {r}^{2}$

where the center is (x_0,y_0)=(-1)-1)

${\left(x + 1\right)}^{2} + {\left(y + 1\right)}^{2} = {5}^{2}$
${x}^{2} + {y}^{2} + 2 x + 2 y + 1 + 1 - 25 = 0$
${x}^{2} + {y}^{2} + 2 x + 2 y - 23 = 0$