How do you find the equation of a circle whose diameter has endpoint (8,0) and (0,6)?

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Answer 1 · 2016-08-30T02:39:09

#x^2+y^2-8x-6y=0#

As the end points of diameter are #(8,0)# and #(0,6)#, the center is their midpoint i.e. #((8+0)/2,(0+6)/2)# or #(4,3)#.

The distance between #(8,0)# and #(0,6)# will be

#sqrt((0-8)^2+(6-0)^2)#

= #sqrt(64+36)#

= #sqrt100=10#

Hence diameter is #10# and radius #10/2=5#.

As the center is #(4,3)# and radius is #5#, the equation of circle is

#(x-4)^2+(y-3)^2=5^2# or

#x^2-8x+16+y^2-6y+9=25# or

#x^2+y^2-8x-6y=0#