Construct two circular arcs meeting at a common point Suppose we are given a ray $\rho_a$ beginning at point $a$ and a ray $\rho_b$ beginning at point $b$. I want to find a circle $C_a$ tangent to $\rho_a$ at point $a$ and another circle $C_b$ tangent to $\rho_b$ at point $b$ such that there exists some point $c$ and a line $\ell$ passing through $c$ such that $C_a$ and $C_b$ intersect at $c$ tangent to $\ell$.
Note that if we insisted on a particular point $c$ and line $\ell$  given ahead of time, then each circle would be over-constrained. However, since we don't insist on a particular point $c$, I think that this should be possible. 


Edit:
Here are two possible solutions (approximate, b/c it's hard to draw)

I'll assume $b$ is not on $\rho_a$.  Start with a circle $C_a$ through $a$ tangent to $\rho_a$ and with $b$ is inside it (any large enough circle through $a$ tangent to $\rho_a$ and on the same side as $b$ will do).  Say $d_a$ is its centre and $r_a$ its radius.  Let $d_b$ be a point on the line through $b$ perpendicular to $\rho_b$ such that $\|d_b - d_a\| + \|d_b - b\| = r_a$.  Then the circle $C_b$ centred at $d_b$ with radius $\|d_b - b\|$ passes through $b$, and $C_a$ and $C_b$ are tangent at the point $c$ where the line through $d_a$ and $d_b$ intersects $C_a$ on the other side of $d_b$.
EDIT: Here's the geometric construction.  Let $L$ be the line through $b$ perpendicular to $\rho_b$.  Let $p$ be on this line at distance $r_a$ from $b$, on the same side of $\rho_b$ as $d_a$.  Then the perpendicular bisector of the line segment from $p$ to  $d_a$ intersects $L$ at $d_b$ which is between
$b$ and $p$ (because $b$ is closer to $d_a$ than it is to $p$), and $\|d_b - d_a\| + \|d_b - b\| = \|p - d_b\| + \|d_b - b\| = r_a$.   

