Given an algebraic plane curve how can one construct a corresponding curve whose degree is one less and such that the points of intersection are also points of tangency?

Specifically, if a plane curve $F$ is given by the polynomial equation $f(x,y) \equiv \sum_{j=0}^{n}\sum_{i=0}^j{a_{i,j}x^{j-i}y^i}=0$, where $x,y \in \mathbb{C}$, how can one construct a corresponding plane curve $\widetilde{F}$, given by $\widetilde{f}(x,y) \equiv \sum_{j=0}^{n-1}\sum_{i=0}^j{\widetilde{a}_{i,j}x^{j-i}y^i}=0$ such that for a pair $p,q$ satisfying

\begin{equation} f(p,q)=\widetilde{f}(p,q)=0 \end{equation}

we also have

\begin{equation} \dfrac{\partial_xf}{\partial_x \widetilde{f}} \bigg|_{x=p,y=q}=\dfrac{\partial_yf}{\partial_y \widetilde{f}} \bigg|_{x=p,y=q} \end{equation}?