How can I generate circle equation from three points using matrices? So, I know how to solve this without using matrices, but I'm not quite sure how it's tackled when using matrices which I need to use. 
I know that the circle has formula of
$ a(x^2+y^2) + bx + cy + d = 0.$
and the circle passes through points
$(-1,0) , (2,3)$ and $ (2,-1).$
I've gotten to the point of creating the matrix
$$
    \begin{bmatrix}
     1 & -1 & 0 & d \\
     13 & 2 & 3 & d \\
     5 & 2 & -1 & d \\
    \end{bmatrix}
$$
but I'm not sure if I need to reduce the matrix or what to carry on. 
 A: $\text{Points: }\{(x_0\mid y_0);(x_1\mid y_1);(x_2\mid y_2)\}$
$\begin{vmatrix}
x^2+y^2&x&y&1\\
x_0^2+y_0^2&x_0&y_0&1\\
x_1^2+y_1^2&x_1&y_1&1\\
x_2^2+y_2^2&x_2&y_2&1\\
\end{vmatrix}=0$
A: You are very close in the creation of your matrix except your last column is a little bit off. Instead of putting $d$ you want to fill it with ones, because the matrix represents the coefficients of the variables. What you are trying to show is:
$$
\begin{bmatrix}
1 & -1 & 0 & 1 \\
13 & 2 & 3 & 1 \\
5 & 2 & -1 & 1\\
\end{bmatrix}
\begin{bmatrix}
a \\ b \\ c \\ d \\
\end{bmatrix}
=
\begin{bmatrix}
0 \\ 0 \\ 0 \\ 0
\end{bmatrix}.
$$
By row reducing the first matrix you will get:
$$
\begin{bmatrix}
1 & 0 & 0 & \frac{1}{3} \\
0 & 1 & 0 & \frac{-2}{3} \\
0 & 0 & 1 & \frac{-2}{3} \\
\end{bmatrix}.
$$
This tells us that $$a = \frac{-1}{3}d,\, b = \frac{2}{3}d,\,  c = \frac{2}{3}d,\, d = \Bbb R \setminus \{0\}.$$
Check it out here by changing the value of d and noticing that the circle will not change.
A: Use matrix equation:
$\begin{bmatrix}
1 &-1 & 0 \\
13 & 2 & 3\\
5 & 2 & -1 \\
\end{bmatrix}\begin{bmatrix}
a \\
b\\
c \\
\end{bmatrix} =\begin{bmatrix}
-d \\
-d\\
-d \\
\end{bmatrix}$ 
You should obtain $a,b,c$ solution as functions of parameter $d$.
