understand the relation between orientation , in circle test and determinant i come from a software engineering background and I'm new to computational geometry ,
I found this picture online and can't manage to understand from where the logic comes from .
I know that determinants are used to know how a transform squishes space but i don't see the link here .

 A: $$\begin{vmatrix}
a_x - c_x & a_y -c_y\\
b_x - c_x & b_y -c_y
\end{vmatrix}$$ is equal to $\Vert ca \Vert \Vert cb \Vert \sin(ca, cb)$ where $\Vert \cdot \Vert$ stands for the norm. The sign of $\sin(ca, cb)$ depends on the fact that $c$ is on one side or the other of the line $ab$. This explains why the sign of the determinant enables to decide on the position of $c$ relatively to the line.
Similar considerations regarding the distance of $d$ vs. the radius of the circle defined by the three points $a,b,c$ enables to explain the second determinant. Have a look to this question.
A: When a continuous curve has the implicit equation of the form $f(x,y)=0$, the sign of the LHS tells you on what side of the curve you are (by continuity, you cannot cross the curve without making the sign change).
Here the determinants enter into play because they form the implicit equations of a line by two points or a circle by three. (Cancelling a determinant expresses lineary dependency.)
$$\begin{vmatrix}x&y&1\\x_0&y_0&1\\x_1&y_1&1\end{vmatrix}=0$$
$$\begin{vmatrix}x^2+y^2&x&y&1\\x_0^2+y_0^2&x_0&y_0&1\\x_1^2+y_1^2&x_1&y_1&1\\x_2^2+y_2^2&x_2&y_2&1\end{vmatrix}=0$$
