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
Stacey Warren - Expert brainly.com
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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
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.
When solving it by matrix, I got c = -5, d= -2, e = 1. But I need know what is wrong with the logic above.
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isn't A, B and C colinear
A (0,1) , B ( 1,3), C ( 2,5)
by elimination ...
d+e = 9
we have a free .... radical? forget the term ... but the matrix is not independant
d+e = 9
d+e = 1
you got a line there
with slope two and y intercept one
must be a magic circle
oh, what @Zarkon said
lol yes they r collinear xD
so circle shuld not exist
slope of AB= slope of BC cx
:) Shame on me.!! Thanks all .
if its a circle, then you are looking at it from the wrong angle :)
But I don't know why I put it on my calculator, it gives me the answer c = -5, d = -2, e =1. :(((
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.