How do you find the equation of the circle with a diameter that has endpoints (-8,0) and (4,-8)?

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Answer 1 · 2016-09-04T15:12:52

Equation of circle is
#x^2+y^2+4x+8y-32=0#

As the end points of diameter are #(-8,0)# and #(4,-8)#, their midpoint is center given by #((-8+4)/2,(0-8)/2)# i.e. #(-2,-4)#.

The distance between #(-8,0)# and #(4,-8)# is given by

#sqrt((4-(-8))^2+(-8-0)^2))#

= #sqrt(144+64)#

= #sqrt208=sqrt(2x2x2×2×13)#

= #4sqrt13#

Hence radius is #2sqrt13#

As center is #(-2,-4)# and radius is #2sqrt13#, the equation of circle is

#(x-(-2))^2+(y-(-4))^2=(2sqrt13)^2# or

#(x+2)^2+(y+4)^2=52# or

#x^2+4x+4+y^2+8y+16=52# or

#x^2+y^2+4x+8y-32=0#