Problem
\[ (x-0)(x-2)+(y-0)(y-3)=0 \] or \[ x^{2}+y^{2}-2 x-3 y=0 \] (c) Three Point Form If three non-collinear points are given, then through them passes a unique circle. In fact, the points are \( \left(x_{1}, y_{1}\right),\left(x_{2}, y_{2}\right) \) and \( \left(x_{3}, y_{3}\right) \) then the equation of the circle will be \[ \left|\begin{array}{llll} x^{2}+y^{2} & x & y & 1 \\ x_{1}^{2}+y_{1}^{2} & x_{1} & y_{1} & 1 \\ x_{2}^{2}+y_{2}^{2} & x_{2} & y_{2} & 1 \\ x_{3}^{2}+y_{3}^{2} & x_{3} & y_{3} & 1 \end{array}\right|=0 \] To see this: suppose that the equation of the circle \[ x^{2}+y^{2}+2 g x+2 f+c=0 \]
Solution
Step 1 ：\begin{vmatrix} x^2+y^2 & x & y & 1 \\ x_{1}^{2}+y_{1}^{2} & x_{1} & y_{1} & 1 \\ x_{2}^{2}+y_{2}^{2} & x_{2} & y_{2} & 1 \\ x_{3}^{2}+y_{3}^{2} & x_{3} & y_{3} & 1 \end{vmatrix}=0
Step 2 ：x^{2}+y^{2}+2gx+2fy+c=0
Step 3 ：2fx+2gy=-x^{2}-y^{2}-c
