Question

any circle can be written in the form \(x^2 + y^2 + cx + dy +e =0\)
Write equation of the circle passes through A (0,1) , B ( 1,3), C ( 2,5). Please, help

Answer 1

hmm a length method will be to find the equation of AB and BC and then their normal
the point where the normals intersect is the centre

once we get the center we can find the radius
and then we r done

Answer 2

Replace:
for A: d + e =-1(1)
for B: c + 3d +e = -10 (2)
for C: 2c + 5d +e = -29 (3)
If I multiple (2) by 2 and subtract (3) from it, I have
d +e = 9
hence with (1) , we have no solution. But it is obviously wrong. However, I don't see what is wrong with the logic.

Answer 3

When solving it by matrix, I got c = -5, d= -2, e = 1. But I need know what is wrong with the logic above.

Answer 4

isn't A, B and C colinear

Answer 5

A (0,1) , B ( 1,3), C ( 2,5)

cx+dy+e=-(x^2+y^2)

0c+1d+1e=-1
1c+3d+1e=-10
2c+5d+1e=-29

by elimination ...

-2c-6d-2e= 20
2c+5d+1e=-29
--------------
d+e = 9

rref{{0,1,1},{1,3,1},{2,5,1}} =

1,0,-2
0,1,1
0,0,0

we have a free .... radical? forget the term ... but the matrix is not independant


d+e = 9
d+e = 1

Answer 6

HI!!

Answer 7

you got a line there
with slope two and y intercept one
must be a magic circle

Answer 8

oh, what @Zarkon said

Answer 9

lol yes they r collinear xD
so circle shuld not exist
slope of AB= slope of BC cx

Answer 10

:) Shame on me.!! Thanks all .

Answer 11

if its a circle, then you are looking at it from the wrong angle :)

Answer 12

But I don't know why I put it on my calculator, it gives me the answer c = -5, d = -2, e =1. :(((

Answer 13

user error?

Answer 14

I put the numbers in 3 x 4 matrix and hit the rref, the result pops out like this. hehehe.... it is crazy like me.

Answer 15

http://www.wolframalpha.com/input/?i=rref {{1%2C3%2C1%2C-10}%2C{2%2C5%2C1%2C-29}%2C{0%2C1%2C1%2C-1}}

Answer 16

firefox likes to post broken links i spose

Answer 17

is it not that