<s>[INST] Why this representation of circle is valid? 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).
 [/INST] 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.
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