Question

How do you write an equation of a circle whose diameter has endpoints (-2, 3) and (4, -1)?

Answer 1

`x^2+y^2-2x-2y-11=0` in the standard form

Explanation:

To obtain the equation of the circle, you need to know its center and radius; the center is the middle point of the given points and the radius is the half of the diameter, so the center is

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

and the radius is

`r=1/2sqrt((4+2)^2+(-1-3)^2)=1/2sqrt(36+16)=1/2sqrt(52)=sqrt(13)`

Then let's substitute them in the equation:

`(x-x_0)^2+(y-y_0)^2=r^2`

to have

`(x-1)^2+(y-1)^2=13`

`x^2+y^2-2x-2y+1+1-13=0`

`x^2+y^2-2x-2y-11=0` in the standard form