How do you write the equation for a circle having (3,0) and (-2,-4) as ends of a diameter?

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Answer 1 · 2016-08-31T03:04:48

Equation of circle is #x^2+y^2-x+4y-6=0#

As the end points of diameter are #(3,0)# and #(-2,-4)#, the center is their midpoint i.e. #((3-2)/2,(0-4)/2)# or #(1/2,-2)#.

The distance between #(3,0)# and #(-2,-4)# will be

#sqrt((3-(-2))^2+(0-(-4))^2)#

= #sqrt(25+16)#

= #sqrt41#

Hence diameter is #sqrt41# and radius #sqrt41/2#.

As the center is #(1/2,-2)# and radius is #sqrt41/2#, the equation of circle is

#(x-1/2)^2+(y+2)^2=(sqrt41/2)^2# or

#x^2-x+1/4+y^2+4y+4=41/4# or

#x^2+y^2-x+4y+4+1/4-41/4=0# or

#x^2+y^2-x+4y+4-10=0# or

#x^2+y^2-x+4y-6=0#