You do have the arbitrary constant term. In the case where $y'' = 0$ you get the solution $y=mx+c$ for two arbitrary constants $m,c$. But you have some extra information, so you can determine $c$. In particular simply plug $y$ back into $(\ast)$ to get $c$ in terms of $m$.

In the other case, you end up with $y^2 = k(x^2-a^2)$ (or something similar) for some arbitrary constant $k$ which yoan then show is infact $-1$ since the distance to the origin is $a^2$.

The lines would necessarily have to meet at some point. But they mean that you need to have two tangent lines on either side of the circle. It's hard to convey this without a drawing - let me know if you're still confused and I'll see what I can do.