Circle determined by $4.89=90-\tan^{-1}(\frac{31.1-x}y)-\tan^{-1}(\frac{y}{36.9-x})$ & $4.89=90-\tan^{-1}(\frac{x-36.9}y)-\tan^{-1}(\frac{y}{x-31.1})$ When graphed, these two equations appear form each half of a circle.
$$
4.89=90-\tan^{-1}\left(\frac{31.1-x}{y}\right)-\tan^{-1}\left(\frac{y}{36.9-x}\right)
$$
$$
4.89=90-\tan^{-1}\left(\frac{x-36.9}{y}\right)-\tan^{-1}\left(\frac{y}{x-31.1}\right)
$$
I am trying to derive the equation for the circle they form, like $\left(x-h\right)^{2}-\left(y-k\right)^{2}=r^{2}$, but have struggling thus far. Through trial and error on Desmos, I managed to determine the following equation:
$$
\left(x-34\right)^{2}+\left(y-33.8761\right)^{2} = 34^{2}\
$$
Its graph appears very similar (within a few d.p.) to that of the two equations above but what I am really interested in is how one would go about deriving such an equation. I understand that there are rather complex distributivity rules for trigonometric  functions, and am wondering what exactly I would need to do to transpose them.
 A: First method:
Group the two "$\tan^{-1}$" into a single one using formula:
$$\tan^{-1}(A)+\tan^{-1}(B)=\tan^{-1}\left(\dfrac{A+B}{1-AB}\right)+R$$
where "residual" $R$ is zero if $AB<1$ and $R=\pm 180°$ otherwise ($180°$ in degrees corresponding to $\pi$ radians, a preferable unit).
(see the graphics here for the understanding of this "$R$")
It will give you for the first formula, after simplification, an equation of the form:
$$\tan^{-1}\left(\dfrac{(a-x)(b-x)+y^2}{((b-x)-(a-x))y}\right)+R=90° - c\tag{1}$$
with $=31.1, b=36.9, c=4.89$.
Taking the tangent on both sides of (1), one gets exactly the equation of a circle:
$$\dfrac{(a-x)(b-x)+y^2}{(b-a)y}=\operatorname{cotan}(c)$$
(I have considered the case $R=0$).
Quick checking: the abscissa of the center is indeed $\frac12(a+b)=34°$.
Second method
Let us give a geometrical interpretation of your first constraint that we write under the form
$$\alpha+\beta=90-4.89$$
where angles $\alpha, \beta$ are such that $\tan \alpha=\dfrac{a-x}{y}$ and $\tan \beta=\dfrac{y}{b-x}$.
Refer now to the following figure where $M(x,y)$ is the current point. Elementary angle chasing show that the angle under which line segment [a,b] is seen from $M$ is $90°-(\alpha+\beta)$, which is a constant. This is a characteristic property of a circle: $M(x,y)$ belongs to a circle passing through points (a,b).
Remark: I am almost sure I have traced back in this way the origin of your problem :).

