Solving the system $(a-b)^2+(c-d)^2=d^2$, $(e-b)^2+(f-d)^2=d^2$ for $b$ and $d$

Question

$$(a-b)^2+(c-d)^2=d^2 \\ (e-b)^2+(f-d)^2=d^2$$

I have this system of two equations, but I have been struggling to isolate the variables, $b$ and $d$. I have tried expanding them, rearranging them, but I really don't get anywhere. Is there something I am missing? Is there another way I can get solutions for this system?

Thank you in advance.

Answer 1

You're basically given points $(a,c),(e,f)$ in Cartesian coordinates and you're asking to find points that have some the same distance $k>0$ from both of them. So you just have to set $(b,d)$ to be some point on the perpendicular bisector of $(a,c),(e,f)$ with a fixed distance from their midpoint. The rest is calculations:

1. The length of the segment is $2L:=\sqrt{(a-e)^2+(c-f)^2}$. Therefore the desired point shall be at a distance of $r:=\sqrt{k^2-L^2}$ of the median.
2. Now note that since the equation of the perpendicular bisector is $\frac{e-a}{c-f}x+v=y$ where $v:=\frac{c+f}2-\frac{a+e}2\cdot\frac{e-a}{c-f}$.
3. Finally note that for the distance to the midpoint is growing on a rate of $u:=\sqrt{\left(\frac{e-a}{c-f}\right)^2+1}$. Hence the reault will be the points: $$[\frac{a+e}2\pm\frac{r}{u},\frac{e-a}{c-f}\left(\frac{a+e}2\pm\frac{r}{u}\right)+v]$$ $$=[\frac{a+e}2\pm\sqrt{\frac{k^2-(\frac{a-e}2)^2-(\frac{c-f}2)^2}{\left(\frac{e-a}{c-f}\right)^2+1}},\frac{e-a}{c-f}\left(\frac{a+e}2\pm\sqrt{\frac{k^2-(\frac{a-e}2)^2-(\frac{c-f}2)^2}{\left(\frac{e-a}{c-f}\right)^2+1}}\right)+\frac{c+f}2-\frac{a+e}2\cdot\frac{e-a}{c-f}]$$