How do you find general form of circle with endpoints of a diameter at (4,3) and (0,1)?

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Answer 1 · 2018-06-14T07:39:25

#x^x+y^2+4x+4y-9=0#

eqn. of circle

#(x-a)^2+(y-b)^2=r^2#

#(a,b)=" the centre"#

#r=" the radius"#

centre is the midpoint of the diameter

#(a,b)=((4+0)/2,(3+1)/2)#

#(a,b)=(2,2)#

#r=sqrt((2-0)^2+(2-1)^2)#

#r=sqrt(4^2+1^2)=sqrt17#

eqn.

#(x-2)^2+(y-2)^2=17#

#x^2+y^2-4x-4y+8=17#

#x^x+y^2-4x-4y-9=0#