Let $g(x,y)=0$ be a closed curve, that means, any point inside that curve satisfies $g(x,y)<0$ and any point outside that curve satisfies $g(x,y)>0$.
Given a point $(a,b)$ outside the curve ($g(a,b)>0$),my question is:
is there one or more points (p,q) satisfying both of the following two equations?
$g(x,y)=0$
$g(a,b)+\frac{\partial g}{\partial x}|_{(a,b)}\cdot(x-a)+\frac{\partial g}{\partial y}|_{(a,b)}\cdot(y-b)=0$
I cannot solve this problem, but I observed that the second of the equation is like the Taylor series: $g(x,y)=g(a,b)+\frac{\partial g(a,b)}{\partial x}\cdot(x-a)+\frac{\partial g(a,b)}{\partial y}\cdot(y-b)+\frac{1}{2}(x-a,y-b)H(s,t)(x-a,y-b)^T$
where $H$ is the Hessian matrix of $g(x,b)$.
Here are two examples which show that such points may or may not exist. Consider the function $$g(x,y):=x^2+\sigma y^2-1\ ,\qquad\sigma\in\{-1,1\}\ ,$$ and let $(a,b):=(1+h,0)$ for a small $h>0$. Plugging this into your equation gives $$(1+h)^2-1 +2(1+h)\bigl(x-(1+h)\bigr)=0\ ,$$ or $$x=1+{2h^2\over 1+h}\ .$$ Now this vertical line does not intersect the circle $x^2+y^2-1=0$, but does intersect the hyperbola $x^2-y^2-1=0$.