an equation of an arc is defined by the equation of a circle:
$$(x-a)^2+(y-b)^2=R^2$$ so $$x=x(y)=a\pm \sqrt{R^2-(y-b)^2}$$ I want to be safe from using the $\pm$ solutions, so I thought if we can say that this arc also belongs to an ellipse of center $(c,d)$, major axis $f-c$ and minor axis $d-g$, so the equation is: $$\left(\dfrac{x-c}{f-c}\right)^2+\left(\dfrac{y-d}{d-g}\right)^2=1$$ so by this i take the positive solution of $x$. I am not sure if this formalism can work.
With some conditions you may assume a part of a circle can be a part of an ellipse too. for this you can construct an ellipse like figure with this method:
1-Take three equal aligned segments AB, BC and CD.
2-Construct equilateral triangle OBC on BC.
3-Draw a circle segment DF center on C and radius CD. Similarly draw a circle segment AE center on B with radius BA.
4- Draw circle segment EF, center on O with radius OE=2 OC=2 OB.
The constructed three center arc is called ellipse-like . As can be seen in figure B and C are the focuses of this figure. Theses are the conditions you must consider for defining the coordinates of centers . Clearly mid point of BC is the center of the shape.