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

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

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?

• Welcome to Math.SE! ... Expanding each equation, the $d^2$ terms cancel. You can solve one equation for the remaining $d$ and substitute into the other (or, if you like, solve both equations for $d$ and set the $d$s equal to each other) to get a solvable quadratic in $b$.
– Blue
Commented Jan 30 at 5:58
• Hello there! I did try that, but when I solved for $d$, I got $d^2=\frac{a^2-2ab+b^2+c^2}{2c}$ which seems like a mess to try and plug into the 2nd equation. Commented Jan 30 at 6:20
• That shouldn't be $d^2$ on the left; just $d$. :) ... You'll get something similar by solving the other equation, which is why setting the $d$s equal to each other may be easier to think about than substituting.
– Blue
Commented Jan 30 at 6:43
• M2 gives $$b^2+\frac{-2ce+2af}{c-f}b+\frac{ce^2-a^2f-c^2f+cf^2}{c-f}=0\\d-\frac{e}{c-f}b+\frac{-a^2-c^2+e^2+f^2}{2(c-f)}=0$$ one way and $$d^2+\frac{-a^2c-c^3+2ace-ce^2-a^2f+c^2f+2aef-e^2f+cf^2-f^3}{c^2-2cf+f^2}d+\frac{a^4+2a^2c^2+c^4-4a^3e-4ac^2e+6a^2e^2+2c^2e^2-4ae^3+e^4+2a^2f^2-2c^2f^2-4aef^2+2e^2f^2+f^4}{4c^2-8cf+4f^2}=0\\b+\frac{c-f}{a-e}d+\frac{-a^2-c^2+e^2+f^2}{2(a-e)}=0$$ the other... Commented Jan 30 at 6:50

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: $$[\frac{a+e}2\pm\frac{r}{u},\frac{e-a}{c-f}\left(\frac{a+e}2\pm\frac{r}{u}\right)+v]$$ $$=[\frac{a+e}2\pm\sqrt{\frac{k^2-(\frac{a-e}2)^2-(\frac{c-f}2)^2}{\left(\frac{e-a}{c-f}\right)^2+1}},\frac{e-a}{c-f}\left(\frac{a+e}2\pm\sqrt{\frac{k^2-(\frac{a-e}2)^2-(\frac{c-f}2)^2}{\left(\frac{e-a}{c-f}\right)^2+1}}\right)+\frac{c+f}2-\frac{a+e}2\cdot\frac{e-a}{c-f}]$$