Let $\Bbb H$ = {$z=x+iy \in \Bbb C | y>0$}. Define $\Bbb H$ lines to be of two kinds, or $L_1$={$x+iy \in \Bbb H |x=c$} or $L_2$= {$x+iy \in \Bbb H | (x-a)^2+y^2=c^2$} with the centre $(a,0)$ on the real axis {$y=0$}.
Show algebraiclly that every pair of distinct points of $\Bbb H$ lies on a unique $\Bbb H$-line.
I can see why this is true, but I'm stuck at proving this algebraically.
Suppose that $(x_1,y_1)$ and $(x_2,y_2)$ are two points. If $x_1=x_2$ then there is a unique line of $L_1$ through these points, otherwise there are no lines of type $L_1$.
If a line $(x-a)^2+y^2=c^2$ passes through our points, then for $a$ and $c$ we get the equations $$(x_1-a)^2+y_1^2=c^2,$$ $$(x_2-a)^2+y_2^2=c^2.$$ It means that $$(x_1-a)^2+y_1^2=(x_2-a)^2+y_2^2,$$ $$x_1^2-2ax_1+a^2+y_1^2=x_2^2-2ax_2+a^2+y_2^2,$$ $$2a(x_2-x_1)=x_2^2-x_1^2+y_2^2-y_1^2.$$ If $x_1=x_2$, then we get $y_2=y_1$, which contradicts that our points are different. Thus in this case we get no lines of type $L_2$ and the line of type $L_1$ through these points is unique.
Otherwise this equation has a unique solution for $a$: $$a=\frac{x_2+x_1}{2}+\frac{y_2^2-y_1^2}{2(x_2-x_1)}.$$ And substituting this $a$ we get the unique $c^2$ - thus, by $c>0$, the unique $c$ - from any of our equations. This means that there is a unique line from $L_2$ (and as we seen, no lines from $L_1$) through our points in this case.