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Recently, I got an equation with this form: $$ (x-x_1)^2 + (y-y_1)^2 = (x-x_2)^2 +(y-y_2)^2 = (x-x_3)^2+(y-y_3)^2$$

Yes, It's from a question about calculating center point from three given points on a circle.

Reduce it, and I got this: $$ \begin{cases} 2(x_2 - x_1)x + 2(y_2-y_1)y = -x_1^2-y_1^2+x_2^2+y_2^2 \\ 2(x_3 - x_1)x + 2(y_3-y_1)y = -x_1^2-y_1^2 +x_2^2+y_2^2 \end{cases} \tag{1} \label{1}$$

Well, it an linear equation, Using Cramer ruler I can get the solution, but The Solution given by my book Taking this Form: $$ x= \frac{ \begin{vmatrix} 1 & x_1^2+y_1^2 & y_1 \\ 1 & x_2^2+y_2^2 & y_2 \\ 1 & x_3^2+y_3^2 & y_3 \\ \end{vmatrix}} {2\begin{vmatrix} 1 & x_1 & y_1 \\ 1 & x_2 & y_2 \\ 1 & x_3 & y_3 \\ \end{vmatrix}}$$ $$ y= \frac{ \begin{vmatrix} 1 & x_1 & x_1^2+y_1^2 \\ 1 & x_2 & x_2^2+y_2^2 \\ 1 & x_3 & x_3^2+y_3^2 \\ \end{vmatrix}} {2\begin{vmatrix} 1 & x_1 & y_1 \\ 1 & x_2 & y_2 \\ 1 & x_3 & y_3 \\ \end{vmatrix}}$$

How this come from? I assume this form are coming from some rewritten of equation $\ref{1}$ . Can someone tell me how to rewrite equation $\ref{1}$ so I can use Cramer ruler and get the solution with the form above?

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  • $\begingroup$ There are typos in the rhs of the second line of $(1)$. $\endgroup$ – Claude Leibovici Apr 14 '17 at 5:47
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$$ \begin{vmatrix} 1 & a & u \\ 1 & b & v \\ 1 & c & w \end{vmatrix}= \begin{vmatrix} 1 & a & u \\ 0 & b-a & v-u \\ 0 & c-a & w-u \end{vmatrix}= \begin{vmatrix} b-a & v-u \\ c-a & w-u \end{vmatrix}$$

The $3\times 3$ determinant looks more elegant than the $2\times 2$ one.

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