If (2,0),(0,1),(4,5),and (0,c) are concyclic then find c
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first find the equation of circle passes through point(2,0) ,(0,1) and (4,5)
equation of the circle
x2 +y2 +2gx +2fy +c =0
so put the points
4+ 4g +c =0 .............1
1 +2f +c =0 ..............2
39 + 8g + 10f +c =0 ............3
solve these equation
g =-25/12
f =-8/3
c =13/3
so equation of circle
x2 + y2 -25x/6 -16y/3 +13/3 =0
now point(0,c) satisfy this equation
c2 -16c/3 + 13/3 =0
3c2 -16c +13 =0
c=1 ,13/3
this is your answer...
equation of the circle
x2 +y2 +2gx +2fy +c =0
so put the points
4+ 4g +c =0 .............1
1 +2f +c =0 ..............2
39 + 8g + 10f +c =0 ............3
solve these equation
g =-25/12
f =-8/3
c =13/3
so equation of circle
x2 + y2 -25x/6 -16y/3 +13/3 =0
now point(0,c) satisfy this equation
c2 -16c/3 + 13/3 =0
3c2 -16c +13 =0
c=1 ,13/3
this is your answer...
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