Contradiction in canonical transformation

The problem I'm supposed to solve is finding $Q$ , such that $(p,q)\rightarrow(P,Q)$ is a canonical transformation. In this case $\mathcal{H}=\frac{p^{2}+q^{2}}{2}$ and the new hamiltonian $\mathcal{K}$ is $\mathcal{K}=P$ . This means $\dot{q}=p$ and $\dot{p}=-q$ Since $\mathcal{H}$ and $\mathcal{K}$ are time independent $\mathcal{H}=\mathcal{K}$ and $P=\frac{p^{2}+q^{2}}{2}$ . Now I use a generating function of canonical transformations $F_{4}=F_{4}(p,P)$ so: $\frac{\partial F_{4}}{\partial p}=-q\quad\quad\quad\mbox{and}\quad\quad\quad\frac{\partial F_{4}}{\partial P}=Q$ $P=\frac{p^{2}+q^{2}}{2}\quad\Rightarrow\quad q=\sqrt{2P-p^{2}}$ Then \begin{equation} F_{4}=-\int\sqrt{2P-p^{2}}dp\quad\Rightarrow\quad Q=-\int \frac{\partial\sqrt{2P-p^{2}}}{\partial P}dp=-arcsin\left(\frac{p}{\sqrt{2P}}\right)=-arcsin\left(\frac{p}{\sqrt{p^{2}+q^{2}}}\right) \end{equation} $\{Q,P\}= \frac{\partial Q}{\partial q}\frac{\partial P}{\partial p}-\frac{\partial Q}{\partial p}\frac{\partial P}{\partial q}=\frac{p}{p^{2}+q^{2}}p-\left(-\frac{q}{p^{2}+q^{2}}\right)q=1$ . Therefore this transformation is canonical. However I also tried to find $Q$ with the generating function $F_{1}=F_{1}(q,Q)$ , where \begin{equation} \frac{\partial F_{1}}{\partial Q}=-P\quad\quad\mbox{and}\quad\quad\frac{\partial F_{1}}{\partial q}=p \end{equation} Then \begin{equation} F_{1}=\int\frac{-p^{2}-q^{2}}{2}dQ\quad\Rightarrow\quad p=\int \frac{\partial\left(\frac{-p^{2}-q^{2}}{2}\right)}{\partial q}dQ=\int -qdQ=-qQ\quad\Rightarrow\quad Q=-\frac{p}{q} \end{equation} This is very different with respect to the first $Q$ found, and $\{Q,P\}=\frac{p}{q^{2}}p+\frac{1}{q}q=\frac{p^{2}}{q^{2}}+1$ which can only be equal to 1 if $p=0$ . But if we assume this is a canonical transformation then $\dot{Q}=1$ and $\dot{P}=0$ , and \begin{equation} \dot{Q}=\frac{\partial Q}{\partial q}\dot{q}+\frac{\partial Q}{\partial p}\dot{p}=\frac{p^{2}}{q^{2}}+1=1\Rightarrow p=0 \end{equation} I think the second result can't be possible, if $p=0$ then $Q=0$ ; so my question is why I could not obtain $Q$ with $F_{1}$ , did I miss something?