Proof of coordinates of centre of conic without partial differentiation.

Question

The centre of the conic represented by:

$S \equiv ax^2+by^2+2hxy+2gx+2fy+c=0$ is represented by $(x,y)\equiv \left(\dfrac{hf-bg}{ab-h^2},\dfrac{gh-af}{ab-h^2}\right)$

I saw it's proof using partial differentiation. Is it possible to prove this without partial differentiation? If yes, how (I do not know partial differentiation therefore I am looking for a different method)?

Answer 1

Let $(u,v)$ be the center of the conic. Let us transfer the origin to $(u,v)$. The new coordinates $(X,Y)$ are given by $X+u = x, Y+v = y$ and the transferred equation is $$a(X+u)^2 + b(Y+v)^2 + 2h(X+u)(Y+v) + 2g(X+u) +2f(Y+v) + c = 0$$ This equation should not have linear terms in $X, Y$. Thus equating the coefficients of $X, Y$ to zero, we get \begin{align*} 2au + 2hv + 2g &= 0\\ 2hu + 2bv + 2f &= 0 \end{align*} Solving for $u, v$, we obtain the given expressions.