Barycentric equation for polar wrt circumcircle

Question

If $P$ is a point on the plane of triangle $ABC$, then we can let $P=(x_0:y_0:z_0)$ be its barycentric coordinates. The barycentric equation of the circumcircle of $ABC$ is $a^2yz+b^2zx+c^2xy=0$. How can I find the equation of $P$'s polar with respect to this circle? In particular, is it $xa(y_0c^2+z_0b^2) + yb(z_0a^2+x_0c^2) + zc(x_0b^2+y^0a^2)$?

Answer 1

Consider the equation of $PA$ and intersect it with the circumcircle $\Gamma$. The solutions are (obviously) $A$ and call the other solution $D$.

Similarly call $E$ the other intersection of $PB$ with $\Gamma$, and $F$ the other intersection of $PC$ with $\Gamma$.

The idea here is to take advantage that the coordinates of the vertices are very simple.

I got $D = (a^2 y_0 z_0 : -y_0 (b^2 z_0 + c^2 y_0) : -z_0 (b^2 z_0 + c^2 y_0))$ for example.

Now $Q = DE\cap AB$ and $R=EF\cap BC$ must lie in the polar. Again, the sides of the triangle have very simple equations.

After some calculations I got the equation of the polar $QR$ to be $$(b^2 z_0 + c^2 y_0) x + (c^2 x_0 + a^2 z_0) y + (a^2 y_0 + b^2 x_0) z = 0$$