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Given the following problem: Finding location of a point on 2D plane, given the distances to three other know points,

we can find the position of the point using 3 other known points, which gives us three equations. My question is, why are 2 points/ 2 equations not enough for these kinds of problems? Two unknowns ($x_0$, $y_0$) should only require two equations, right?

More precisely, given $x_1$, $y_1$, $x_2$, $y_2$, $d_1$ and $d_2$, why can't we solve for $x_0$ and $y_0$ (exactly) using the following formulas:

$$(x_0-x_1)^2+(y_0-y_1)^2=d_1^2$$ $$(x_0-x_2)^2+(y_0-y_2)^2=d_2^2 $$

I understand that the squares introduce $\pm$, but I've never needed more than 2 equations to solve for 2 unknowns before and it's rubbing me the wrong way.

Thank you!

Bouch
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    Hint: each equation represents a circle. Two circles can intersect at 0,1 or 2 points. – DatBoi Jan 08 '23 at 05:59
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    @DatBoi or every point – David Raveh Jan 08 '23 at 06:10
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    @DavidRaveh ah yes of course! I believe the given points would be distinct, so that shouldn't be a problem. – DatBoi Jan 08 '23 at 06:23
  • @DatBoi The intuition for it does make sense, but having never needed 3 equations for 2 unknowns, I am still quite uncomfortable with this. Does this only happen when the variables we are solving for are squared? Or can this happen in other circumstances?

    Thank you!|

     – Bouch Jan 09 '23 at 00:13
  • @Bouch You can absolutely solve for the unknowns using just two equations. Solving this may yield, for say, 2 sets of solutions, like i mentioned above. However, notice that we're trying to determine a unique point in the plane. Adding another equation possibly gets rid of the other solution, yielding you a unique solution, ie, a unique point in the plane. – DatBoi Jan 09 '23 at 06:32
  • @DatBoi Hmmm, sounds fair. I guess each equation reduces the size of the solution space by 1, and 2 solutions are not equivalent to a whole degree of freedom. Thank you! – Bouch Jan 12 '23 at 01:08

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