Finding coordinate equations for graphing of problem. Have a request to help me in draw using desmos the "Treasure hunt problem" (explained in detail with basic approaches here with desired implementation as in basic option given here. 
I expect that the best approach is to have two circles tangentially intersecting each other, and their centers representing 'Pine' and 'Oak' trees respectively; while the intersection of two circles being the location of 'gallows'. Also, the each circles' two radius are perpendicular to each other. The other end on the perpendicular radii represents 'Spike1', 'Spike2' respectively.
I  request some help in finding the intersection point of two circles that represents the gallows, as the main issue is this configuration is not workable , i.e. the intersection of two circles is meaningless once the relative angle among the pine and oak trees (the two radii) is changed.

The only way is for the two circles to be intersecting still, but be able to to move while intersecting at more than one point possibly, as in the below diagram. 
Anyway, the approach should ensure the independence of treasure from the position of the gallows, as shown in the first link. also, may be for the above approach using circles, polar coordinates' equations may work to represent using coordinate geometry in desmos.
 Would prefer the circle based approach, as above.
An alternate approach as shown in the first link also, is to place the trees on the x-axis, with gallows changing location along the locus of some parametric equation as a function of distance from the trees.
In the first approach, the movement of gallows is not possible, and only a particular configuration can be shown like in the first / second diagrams, while in the second approach (as shown in the third diagram, below) the gallows moves as a function on locus.
The second approach is shown below:
 A: Still not sure what you are after, though, but I have constructed the model using functions in Desmos

So you have circles defined by:
$$\begin{align}\left(x-h_1\right)^2+\left(y-k_1\right)^2=r_1^2\tag{1}\\
\left(x-h_2\right)^2+\left(y-k_2\right)^2=r_2^2\tag{2}\end{align}$$
Where $(1)$'s center is the Oak$(h_1,k_1)$ and $(2)$'s center is the Pine$(h_2,k_2)$.
The location of the Gallows is defined by $(G_x,G_y)$, of course the radii of $(1)$ and $(2)$ are defined by:
$$r_1=\sqrt{\left(G_x-h_1\right)^2+\left(G_y-k_1\right)^2}\\r_2=\sqrt{\left(G_x-h_2\right)^2+\left(G_y-k_2\right)^2}\\
$$
The lines connecting Oak and Pine to Gallows are defined by:
$$y-k_1=\left(M_1\right)\left(x-h_1\right)\\y-k_2=\left(M_2\right)\left(x-h_2\right)\\
$$
Where $M_1$ and $M_2$ are the slopes of the respective lines. Of course, to get the $90^\circ$ turn, the perpendicular lines equations will be the same as above except the slopes are the negative reciprocal, thus:
$$\begin{align}
y-k_1&=-\left(M_1\right)^{-1}\left(x-h_1\right)\tag{3}\\
y-k_2&=-\left(M_2\right)^{\left(-1\right)}\left(x-h_2\right)\tag{4}\\
\end{align}$$
To find the Spikes, you just have to find the intersection between $(1)$ & $(3)$, and $(2)$ & $(3)$, thus you get:
$$\begin{align}\text{$(1)$ & $(3)$}\iff (K_x,K_y)&\Rightarrow K_x=\frac{h_1M_1^2+h_1+\sqrt{M_1^2+1}M_1r_1}{M_1^2+1}\\
&\Rightarrow K_y=\frac{k_1M_1^2+k_1-\sqrt{M_1^2+1}r_1}{M_1^2+1}\\
\text{$(2)$ & $(4)$}\iff (P_x,P_y)&\Rightarrow P_x=\frac{h_2M_2^2+h_2+\sqrt{M_2^2+1}M_2r_2}{M_2^2+1}\\
&\Rightarrow P_y=\frac{k_2M_2^2+k_2-\sqrt{M_2^2+1}r_2}{M_2^2+1}
\end{align}$$
Thus, simply, the Treasure is the midpoint between $(K_x,K_y)$ and $(P_x,P_y)$, which should be:
$$M_{P,K}=\left(\frac{P_x+K_x}{2},\frac{P_y+K_y}{2}\right)$$

If you want the two circles tangent at the Gallows$(G_x,G_y)$, then you have to make the Oak, Pine and Gallows collinear, say by lying in the $x$-axis. You can freely move the Gallows then across the $x$-axis. And as mentioned in the first link, the Treasure is invariant to the location of the Gallows.


A: If we put the Oak tree at A, and the Pine tree at B, and your starting point at X.  
And we plot it in the complex plane.  
Then spike 1 is at $A + i(A-X)$ 
And spike 2 is at $B - i(A-X)$
Putting the treasure at $\frac {A+B}{2} + \frac {A-B}{2} i$
And clearly the location of the treasure does not depend on $X.$
