Inferring behavior of a function on the plane from curve integral

Let $P,Q:\mathbb{R}^2\rightarrow\mathbb{R}$ be two $C^1$ functions on the plane. Denote by $\Gamma$ the unit circle. $(P^2+Q^2)|_\Gamma>0$ and $\oint_\Gamma \frac{PdQ-QdP}{P^2+Q^2}\neq0$. Prove there exist a point $(x_1,y_1)$ on $\mathbb{R}^2$ such that $P(x_1,y_1)=Q(x_1,y_1)=0$.

I tried to use proof by contradiction and use the Green theorem. I found that when $P\neq0$ ,$d\arctan(\frac{Q}{P})=\frac{PdQ-QdP}{P^2+Q^2}$. But I cannot handle the points on which $P=0$. Can someone offer any help? Thanks.

Assume by contradiction that there is no point $(x_1,y_1)$ on $\mathbb{R}^2$ such that $$P(x_1,y_1)=Q(x_1,y_1)=0 \quad \Leftrightarrow\quad P^2(x_1,y_1)+Q^2(x_1,y_1)=0.$$ Then the continuous function $P^2+Q^2$ has a positive minimum in the closed unit disk $D$, and we can apply Green's Theorem with respect to $D$, getting a contradiction $$\begin{align}0\not=\oint_\Gamma \frac{PdQ-QdP}{P^2+Q^2}&=\oint_\Gamma \frac{PQ_x-P_xQ}{P^2+Q^2}dx+\frac{PQ_y-P_yQ}{P^2+Q^2}dy\\&=\iint_D\left(\frac{\partial}{\partial x}\left(\frac{(PQ_y-P_yQ)}{P^2+Q^2}\right)-\frac{\partial}{\partial y}\left(\frac{(PQ_x-P_xQ)}{P^2+Q^2}\right)\right)dxdy=0 \end{align}$$ because $\frac{\partial}{\partial x}\left(\frac{PQ_y-P_yQ}{P^2+Q^2}\right)-\frac{\partial}{\partial y}\left(\frac{PQ_x-P_xQ}{P^2+Q^2}\right)$ is identically zero inside $D$. Actually, in this way, we show that there exists such point $(x_1,y_1)$ inside $D$.