Why this representation of circle is valid?

Question

A line passing through two distinct points $P_1(x_1,y_1),P_2(x_2,y_2)$ can be expressed by $$\det\left| \begin{array}{ccc} x-x_1&y-y_1 \\ x_2-x_1&y_2-y_1 \\ \end{array} \right|=0$$

Since line is set $A=\{a P_1+bP_2|a+b=1\}$ and it corresponds to determinant properties.

A circle passing three(non-colinear) points $(x_1,y_1),(x_2,y_2),(x_3,y_3)$ can be expressed by $$\det\left| \begin{array}{ccc} (x-x_1)^2+(y-y_1)^2 & (x-x_1) & (y-y_1) \\ (x_2-x_1)^2+(y_2-y_1)^2 & (x_2-x_1) & (y_2-y_1) \\ (x_3-x_1)^2+(y_3-y_1)^2 & (x_3-x_1) & (y_3-y_1) \end{array} \right|=0$$ or $$\det\left| \begin{array}{ccc} x^2+y^2 & x & y&1 \\ x_1^2+y_1^2 & x_1 & y_1&1 \\ x_2^2+y_2^2 & x_2 & y_2&1 \\ x_3^2+y_3^2 & x_3 & y_3&1 \\ \end{array} \right|=0$$

Please explain simple and neat as possible(like in determinant sense).

Answer 1

Consier the set of equations $$Ax^2+Ay^2+Bx+Cy+D=0 \tag{1} $$ $$Ax_1^2+Ay_1^2+Bx_1+Cy_1+D=0 \tag{2} $$ $$Ax_2^2+Ay_2^2+Bx_2+Cy_2+D=0 \tag{3} $$ $$Ax_3^2+Ay_3^2+Bx_3+Cy_3+D=0 \tag{4} $$ this is a $4\times4$ linear system and it must satisfy, $$\left( \begin{array}{ccc} x^2+y^2 & x & y&1 \\ x_1^2+y_1^2 & x_1 & y_1&1 \\ x_2^2+y_2^2 & x_2 & y_2&1 \\ x_3^2+y_3^2 & x_3 & y_3&1 \\ \end{array} \right) \left( \begin{array}{ccc} A\\ B \\ C \\ D \\ \end{array} \right)=0 \tag{5}$$ taking determinant you get your required condition. Also you can subtract the set of equations $1\to 4$ to reduce it into $3\times 3$ system and get the other result.

Answer 2

Hint for the $4\times4$ version.

• Explain why expanding the determinant gives an equation of the form $$Ax^2+Ay^2+Bx+Cy+D=0\ ,$$ where $A,B,C,D$ are constants.
• Explain why $A\ne0$.
• Explain why the three given points satisfy the equation.

Good luck!