I am working within the upper half plane.
I broke this question into three categories:
- $\ell_1$ is a vertical line.
- $\ell_1$ is a semi-circle and $p$ is directly above the center of $\ell_1$.
- $\ell_1$ is a semi-circle and $p$ is not directly above the center of $\ell_1$.
I have proved the first two cases, but I am now stuck on the last case.
For case three it is clear that $\ell_2$ must be a also be a semi-circle. Since both lines are centered on $\mathbb{R}$, we can tell that they intersect at $\pi/2$ rad if and only if $|c_1-c_2|^2+r_2^2=r_1^2$. Where $|c_1-c_2|$ is the distance between both lines centers and $r_i$ corresponds to the radius of the respective line.
We also know that $p\in\ell_2 \implies r_2 = \sqrt{(Re(p)-c_2)^2+Im(p)^2}$. Now substituting the second equation into the first gives us the following: $$2c_2^2-2c_1c_2+c_2^2+Re(p)^2-2c_2Re(p)+c^2+Im(p)^2=r_1^2$$ Isolating for $c_2$ using the quadratic equation gives us the following: $$c_2 = \frac{2(c_1+Re(p)) \pm \sqrt{8r_1^2-4c_1^2+8c_1Re(p)-8Im(p)^2-4Re(p)^2}}{4}$$ If the statment that im trying to prove is always true, we should see that $8r_1^2-4c_1^2+8c_1Re(p)-8Im(p)^2-4Re(p)^2 = 0$ for all cases. However considering $c_1=r_1=1, Re(p) = 2, \Im(10)$ we see that $8r_1^2-4c_1^2+8c_1Re(p)-8Im(p)^2-4Re(p)^2<0$.
This leads me to believe that the question is flawed as I have just demonstrated that when $\ell_1$ is a semi-circle with center at 1 and radius 1, there does not not exist a another semi-circle such that it can intersect $\ell_2$ and hit the point $p=2 + 10i$.
I would appreciate if someone could verify my proof and/or point out any flaws in it.