Geo4.15

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Find the equation to the circle passing through the points(1,3),(0,-2),(-3,1)\,


Let the equation of the circle be

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

The point (1,3) lies on the circle,then

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

2g+6f+c=-8\,

Let this equation be 1.

The point (0,-2) lies on the circle,then

0^{2}+(-2)^{2}+2g(0)+2f(-2)+c=0\,

-4f+c=-4\,

Let this equation be 2.

The point (-3,1) lies on the circle,then

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

-6g+2f+c=-10\,

Let this equation be 3.

Solving 1 and 3 we get

20f+4c=-34\,

Solving this and equation 3, we get

9c=-54\,

c=-6\,

Hence,

f={\frac  {-1}{2}},-6g-1-6=-10\,

g={\frac  {1}{2}},f={\frac  {-1}{2}},c=-6\,

Now the equation of the circle is

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


Main page:Geometry:Circles