# What is the name of this arbelos-like shape?

What is the name for the shape enclosed by three tangential arcs of circles as shown in the diagram C F C''? If the three circles' centres (O, E & E'') were co-linear, it would be an Arbelos, but they are not.

Further, does anyone have any idea how to find the inscribed circle on OA that is tangential to all three arcs? (COC'' is a sector of an octagon, but I'd like a general method for other polygons).

The special case of the theorem states that for four mutually tangent circles, their signed curvatures satisfy the relationship $$(\kappa_1 + \kappa_2 + \kappa_3 + \kappa_4)^2 = 2(\kappa_1^2 + \kappa_2^2 + \kappa_3^2 + \kappa_4^2).$$ here, $|\kappa_i| = 1/r_i$, where $r_i$ is the radius of each circle, and the sign of $\kappa_i$ is negative if the tangency is internal for that circle--in your case, the arc $C''C$ with center $O$ is a portion of an internally tangent circle with radius $OA$, thus its curvature is the negative reciprocal of $OA$. Since you know the radii of three of the four circles in question, the desired fourth circle (which is externally tangent) will have a positive curvature and its inradius can be found by solving the resulting quadratic.
Once the four signed curvatures are known by solving the above, the incenter of the fourth circle can be calculated using the generalized version of the theorem, which is best conceptualized by placing the figure on the complex plane. In particular, if $z_i$ represents the complex-valued coordinate of the $i^{\rm th}$ circle, then $$(\kappa_1 z_1 + \kappa_2 z_2 + \kappa_3 z_3 + \kappa_4 z_4)^2 = 2(\kappa_1^2 z_1^2 + \kappa_2^2 z_2^2 + \kappa_3^2 z_3^2 + \kappa_4^2 z_4^2).$$ Note this equation is invariant to translation or rotation in the plane, so for instance, if $\kappa_1$ is the curvature of circle $O$, we may take without loss of generality $z_1 = 0 + 0i$; i.e., place $O$ at the origin. Then the incenter of the desired circle is unique up to some rotation about $O$.