Determinant formula for coordinates of circumcenter and orthocenter of a triangle

I've come across the following formulas for coordinates of circumcenter $O=(x_O,y_O)$ and orthocenter $H=(x_H,y_H)$ of a triangle, in a formula book, stated without derivation. For a triangle with vertices $(x_i,y_i),$ $i \in \{1,2,3\}$,

$$x_O=\frac{\begin{vmatrix} x_1^2+y_1^2 & y_1 & 1 \\ x_2^2+y_2^2 & y_2 & 1 \\ x_3^2+y_3^2 & y_3 & 1 \end{vmatrix}}{2\begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix}} \quad , \quad y_O=\frac{\begin{vmatrix} x_1 & x_1^2+y_1^2 & 1 \\ x_2 & x_2^2+y_2^2 & 1 \\ x_3 & x_3^2+y_3^2 & 1 \end{vmatrix}}{2\begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix}} \tag{A}$$

$$x_H=\frac{\begin{vmatrix} y_1 & x_2x_3+y_1^2 & 1 \\ y_2 & x_3x_1+y_2^2 & 1 \\ y_3 & x_1x_2+y_3^2 & 1 \end{vmatrix}}{\begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix}} \quad , \quad y_H=\frac{\begin{vmatrix} x_1^2+y_2y_3 & x_1 & 1 \\ x_2^2+y_3y_1 & x_2 & 1 \\ x_3^2+y_1y_2 & x_3 & 1 \end{vmatrix}}{\begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix}} \tag{B}$$

The circumcenter case is pretty clear to me. Comparing two different equations of circumcircle $$\begin{vmatrix} x^2+y^2 & x & y & 1 \\ x_1^2+y_1^2 & x_1 & y_1 & 1 \\ x_2^2+y_2^2 & x_2 & y_2 & 1 \\ x_3^2+y_3^2 & x_3 & y_3 & 1 \end{vmatrix} = 0 = \lambda(x^2+y^2 -2x_{O}x -2y_{O}y+c)$$

and equating the coefficients of $x,y$, one obtains $(A)$.

But I have not been able to make progress towards derivation of $(B)$. Can somebody provide a proof of it?

Please avoid an algebra-tedious proof. Thank you!

Solution 1:

Each point verifies \begin{eqnarray*} (x_k-x_O)^2 + (y_k-y_O)^2 &=& r^2 \\ x_k^2+y_k^2+ x_O^2+y_O^2- 2(x_kx_O+y_ky_O) &=& r^2 \\ 2(x_kx_O+y_ky_O) + (r^2-x_O^2-y_O^2) &=& x_k^2+y_k^2 \end{eqnarray*}

This yields to the linear system $$ \begin{pmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{pmatrix} \begin{pmatrix} 2x_O \\ 2y_O \\ a \end{pmatrix} = \begin{pmatrix} x_1^2+y_1^2 \\ x_2^2+y_2^2 \\ x_3^2+y_3^2 \end{pmatrix} $$ It can be solved using Cramer's rule. which yields to (A).

Note $D$ the denominator.

For the orthocenter, we use the relation \begin{eqnarray*} x_H-x_O &=& (x_1-x_O)+(x_2-x_O)+(x_3-x_O) \\ x_H &=& (x_1+x_2+x_3)-2 x_O \\ D x_H &=& (x_1+x_2+x_3)D+ \begin{vmatrix} y_1 & x_1^2+y_1^2 & 1 \\ y_2 & x_2^2+y_2^2 & 1 \\ y_3 & x_3^2+y_3^2 & 1 \end{vmatrix} \\ D x_H &=& \begin{vmatrix} y_1 & -(x_1+x_2+x_3)x_1 & 1 \\ y_2 & -(x_1+x_2+x_3)x_2 & 1 \\ y_3 & -(x_1+x_2+x_3)x_3 & 1 \end{vmatrix} + \begin{vmatrix} y_1 & x_1^2+y_1^2 & 1 \\ y_2 & x_2^2+y_2^2 & 1 \\ y_3 & x_3^2+y_3^2 & 1 \end{vmatrix} \end{eqnarray*} Doing computations yield (B).

Solution 2:

Consider instead triangle $A_1B_1C_1$ whose midtriangle is precisely triangle $ABC$.

A classical result says that the orthocenter $H$ of $ABC$ is the circumcenter of $A_1B_1C_1$.

It is easy to express the coordinates of points $A_1,B_1,C_1$ as functions of the coordinates of $A,B,C$ using relationships

$$\vec{AB}+\vec{AC}=\vec{AA_1} \implies A_1=B+C-A, \ \text{etc.} $$

Last step: one applies the first set of formulas to triangle with vertices $A_1,B_1,C_1$ .

Please note that my first explanation using barycentric coordinates was a much too indirect way.