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.

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

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

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