Find center of rational circle When a circle is defined by three points on its perimeter, it's possible to find its center $(h,k)$ using:
$$
h=\frac{\left| \begin{matrix}x_1^2+y_1^2 & y_1 & 1 \\x_2^2+y_2^2 & y_2 & 1 \\ x_3^2+y_3^2 & y_3 & 1\end{matrix} \right|}{2*\left| \begin{matrix}x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{matrix} \right|},
k=\frac{\left| \begin{matrix}x_1 & x_1^2+y_1^2  & 1 \\x_2 & x_2^2+y_2^2 & 1 \\ x_3 & x_3^2+y_3^2 & 1\end{matrix} \right|}{2*\left| \begin{matrix}x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{matrix} \right|}
$$
(Taken from here)
But this assumes that the points are defined using real numbers. 
How should this formula be adapted when the points are defined using rationals with a common denominator? (i.e.: a point is defined using $(a/w, b/w)$).
I noticed that the denominator is the same as when calculating the orientation of three points. In that situation the 1's are replaced by the $w$'s. Is this also true for finding the center? Since the numerator contains squares, what are the implications then for the last column?
Update:
Based on @saulspatz answer I think the following is correct:
My actual situation is as follows (only for $h$):
$$
h=\frac{\left| \begin{matrix}\frac{x_1}{w_1}^2+\frac{y_1}{w_1}^2 & \frac{y_1}{w_1} & 1 \\\frac{x_2}{w_2}^2+\frac{y_2}{w_2}^2 & \frac{y_2}{w_2} & 1 \\ \frac{x_3}{w_3}^2+\frac{y_3}{w_3}^2 & \frac{y_3}{w_3} & 1\end{matrix} \right|}{2*\left| \begin{matrix}\frac{x_1}{w_1} & \frac{y_1}{w_1} & 1 \\ \frac{x_2}{w_2} & \frac{y_2}{w_2} & 1 \\ \frac{x_3}{w_3} & \frac{y_3}{w_3} & 1 \end{matrix} \right|}
$$
When I now multiply each row by its corresponding $w$ I have to correct the determinant by dividing by the same factor:
$$
h=\frac{\frac{1}{w_1^2w_2^2w_3^2}\left| \begin{matrix}x_1^2+y_1^2 & y_1w_1 & w_1^2 \\x_2^2+y_2^2 & y_2w_2 & w_2^2 \\ x_3^2+y_3^2 & y_3w_3 & w_3^2\end{matrix} \right|}{\frac{2}{w_1w_2w_3}\left| \begin{matrix}x_1 & y_1 & w_1 \\ x_2 & y_2 & w_2 \\ x_3 & y_3 & w_3 \end{matrix} \right|}
$$
I now multiply by $w_1^2w_2^2w_3^2$, which leaves:
$$
h=\frac{\left| \begin{matrix}x_1^2+y_1^2 & y_1w_1 & w_1^2 \\x_2^2+y_2^2 & y_2w_2 & w_2^2 \\ x_3^2+y_3^2 & y_3w_3 & w_3^2\end{matrix} \right|}{2w_1w_2w_3\left| \begin{matrix}x_1 & y_1 & w_1 \\ x_2 & y_2 & w_2 \\ x_3 & y_3 & w_3 \end{matrix} \right|}
$$
 A: If you multiply a column of a determinant by $c$ the determinant is also multiplied by $c$.  So if we say that the points are $(\frac{x_k}{w}, \frac{y_k}{w})$ the values would become $$
h=\frac{\frac{1}{w^3} \left|\begin{matrix}x_1^2+y_1^2 & y_1 & 1 \\x_2^2+y_2^2 & y_2 & 1 \\ x_3^2+y_3^2 & y_3 & 1\end{matrix} \right|}{\frac{2}{w^2}\left| \begin{matrix}x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{matrix} \right|},
k=\frac{\frac{1}{w^3}\left| \begin{matrix}x_1 & x_1^2+y_1^2  & 1 \\x_2 & x_2^2+y_2^2 & 1 \\ x_3 & x_3^2+y_3^2 & 1\end{matrix} \right|}{\frac{2}{w^2}\left| \begin{matrix}x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{matrix} \right|}$$
Now after multiplying numerator and denominator by $w^3$, you could bring the the $w$ inside the determinant by multiplying the last column by $w$.  That would make the last column $w$, but only in the denominators.  
A: The calculation/formula is exactly the same for rational values of the entries.
Your profile suggests that you want to do this in a computer program. To get an exact rational answer rather than a floating point approximation you will have to do rational number arithmetic. Some languages implement this for you.
The squares don't cause a problem - they are rational. If you needed square roots you'd be in trouble.
