How can I generate circle equation from three points using matrices?

Question

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.

Answer 1

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.

Answer 2

$\text{Points: }\{(x_0\mid y_0);(x_1\mid y_1);(x_2\mid y_2)\}$
$\begin{vmatrix} x^2+y^2&x&y&1\\ x_0^2+y_0^2&x_0&y_0&1\\ x_1^2+y_1^2&x_1&y_1&1\\ x_2^2+y_2^2&x_2&y_2&1\\ \end{vmatrix}=0$

Answer 3

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 $d$.