I have to prove that if P is a R-module

What is the easier way to find the circle given three points?
Given three points ${\displaystyle (x_1,y_1),(x_2,y_2)}$, and ${\displaystyle (x_3,y_3)}$, if
${\displaystyle \frac{y_2-y_1}{x_2-x_1}\ne \frac{y_3-y_2}{x_3-x_2}\ne \frac{y_1-y_3}{x_1-x_3},}$
then there will be a circle passing through them. The general form of the circle is
${\displaystyle x^2+y^2+ dx +ey+f=0.}$
By substituting ${\displaystyle x=x_i\text{and}y=y_i}$, there will be a system of equation in three variables, that is:
$\begin{array}{rl}\left(\begin{array}{ccc}{x}_1& y_1& 1\\ x_2& y_2& 1\\ x_3& y_3& 1\end{array}\right)\left(\begin{array}{c}d\\ e\\ f\end{array}\right)& =\left(\begin{array}{c}-(x_1^2+y_1^2)\\ -(x_2^2+y_2^2)\\ -(x_3^2+y_3^2)\end{array}\right).\end{array}$
As there are a lot of things going around, the solution is prone to errors. Maybe this solution also has an error.
Is there a better way to solve for the equation of the circle?

Reduce the system
(D2 + 1)[x] − 2D[y] = 2t
(2D − 1)[x] + (D − 2)[y] = 7.
to an equivalent triangular system of the form
P1(D)[y] = f1(t)
P2(D)[x] + P3(D)[y] = f2(t)
and solve.

Find Laplace transform of L[te
t
sin4t] is

Suppose$U=\{(x,y,x+y,x-y,2x)\in F^5,x,y\in F\}$ . Find a subspace $W$of $F^5$ such that $F^5=U\oplus W$.

In the relation in the table below, write a value that will make the relation not   represent a function.