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

148 Points
13 years ago

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

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Keshav Jhawar
18 Points
10 years ago

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.

ankit singh
3 years ago
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
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