How do you write the equation for a circle endpoints of a diameter at (3,8) and (-7,4)?

1 Answer
May 16, 2016

Answer:

The equation of circle is #x^2+y^2+4x-12y+11=0#

Explanation:

The center of the circle is mid point between #(3,8)# and #(-7,4)# i.e. #((3-7)/2,(8+4)/2# or #(-2,6)#.

The radius is distance between #(-7,4)# and #(-2,6)# or #sqrt((-2+7)^2+(6-4)^2)=sqrt(25+4)=sqrt29#

Hence equation of circle of radius #sqrt29# and center #(-2,6)# is

#(x+2)^2+(y-6)^2=29# or

#x^2+4x+4+y^2-12y+36-29=0# or

#x^2+y^2+4x-12y+11=0#