Geo4.28

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Find the equation of the circle which passes through (1,1),(2,2) and whose radius is unity.

Let the equation be

x^{2}+y^{2}+2gx+2fy+c=0\,

The circle passes through (1,1)

Hence

2g+2f+c+2=0\,

Let this be 2

It passes through (2,2)

Now

4g+4f+c+8=0\,

Let this be 3

Radius is

g^{2}+f^{2}-c=1\,

Let this be 4.

Subtracting 2 from 3, we get

g+f+3=0\,

This is equation 5.

Adding 2 and 4 equations we get

g^{2}+f^{2}+2g+2f+1=0\,

Substituting f=-g-3 from 4,

g^{2}+(g+3)^{2}+2g-2(g+3)+1=0\,

<amth>g^2+3g+2=0,g=-1,-2\,</math>

Hence

f=-2,-1\,

The value of c is 4,4

Hence the equation is

x^{2}+y^{2}-2x-4y+4=0,x^{2}+y^{2}-4x-2y+4=0\,


Main Page:Geometry:Circles