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

1 Answer
Aug 23, 2016

x^2+y^2-2x-2y-11=0x2+y22x2y11=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)M(2+42;312)=(1;1)

and the radius is

r=1/2sqrt((4+2)^2+(-1-3)^2)=1/2sqrt(36+16)=1/2sqrt(52)=sqrt(13)r=12(4+2)2+(13)2=1236+16=1252=13

Then let's substitute them in the equation:

(x-x_0)^2+(y-y_0)^2=r^2(xx0)2+(yy0)2=r2

to have

(x-1)^2+(y-1)^2=13(x1)2+(y1)2=13

x^2+y^2-2x-2y+1+1-13=0x2+y22x2y+1+113=0

x^2+y^2-2x-2y-11=0x2+y22x2y11=0 in the standard form