Find a point $P$ on the line $_{-3}L_{\sqrt{7}}$ in the Poincare plane whose coordinate (ruler) is $2$.
Let $P =(x,y)$. The line is on the Poincare plane, so it is a semicircle on the upper half-plane with center $-3$ and radius $\sqrt{7}$. The ruler equation for a Poincare plane is $\ln(\frac{x-c+r}{y})$ which equals $2$. Thus $\ln(\frac{x-c+r}{y}) = 2$.
Is my reasoning correct? How can I find point $P$?
In $\mathscr{H}$, we know that $$_c\mathcal{L}_r=\{(x,y)\in\mathscr{H}\mid(x-c)^2+y^2=r^2\}$$ Considering the position of $P(x_p,y_p)$ we get $$(x_p+3)^2+y_p^2=7~~~~~~~~~~~~~(I)$$ But $f(x_p,y_p)=2$ where $$f(x,y)=\ln\left(\frac{x-c+r}{y}\right)$$ so $\ln\left(\frac{x_p+3+\sqrt{7}}{y_p}\right)=2$ and so $$(x_p+3)^2=(e^2y_p-\sqrt{7})^2~~~~~~~~~~~~~(II)$$ Now solve (I) and (II) simultaneously.