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

1 Answer
Aug 11, 2016

Answer:

#x^2+y^2+2x+2y-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((x_2-x_1)^2+(y_2-y_1)^2)#

#d=sqrt((3+5)^2+(-4-2)^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:

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

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

and the radius is 5:

#(x+1)^2+(y+1)^2=5^2#

#x^2+y^2+2x+2y+1+1-25=0#

in the standard form:

#x^2+y^2+2x+2y-23=0#