If the points (2,0),(0,1),(4,5)and (0,c)are concyclic, then the value of \"c\" is ?

Dear tadiparthi

first find the equation of circle passes through point(2,0) ,(0,1) and (4,5)

equation of the circle

x² +y_² +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

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Badiuddin

The third eq is most probably wrong as 5^2 + 4^2 is 41 and not 39 as you have mentioned. If there is any fault in what I said please correct me.