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

1 Answer
Gerardina C.
Aug 23, 2016

#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

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