Geo4.17

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Show that the points (-6,0),(-2,2),(-2,-8),(1,1)\, are concyclic


Let the equation of the circle be

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

The point (-6,0) lies on the circle,then

36-12g+c=0\,

-12g+c=-36\,

Let this equation be 1.

The point (-2,2) lies on the circle,then

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

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

Let this equation be 2.

The point (-2,-8) lies on the circle,then

4+64-4g-16f+c=0\,

-4g-16f+c=-68\,

Let this equation be 3.

Solving 1 and 3 we get

g-2f=-4\,

Solving this and equation 2, we get

c=-12\,

Hence,

g=2\,

f=3,g=2,c=-12\,

Now the equation of the circle is

x^{2}+y^{2}+4x+6y-12=0\,

If the fourth point (1,1)lies on this circle,then the four points are concyclic.

Now

1+1+4+6-12=0\,


Main page:Geometry:Circles