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This problem seems so simple that it must have been answered before, but I couldn't figure out what to search. Anyway,

Suppose you have the Euclidean distance between two points, $a$ and $b$, $z = \sqrt{x^2+y^2}$, where $x$ and $y$ are the differences, respectively, between the $x$ and $y$ co-ordinates of $a$ and $b$. Note that you do NOT know $a$ and $b$, only $z$. How do you find the value of $z' = \sqrt{(x-1)^2+(y-1)^2}$?

EDIT: Some people have (correctly) pointed out that I seem to be talking about distance from the origin. I wrote the original post in a hurry and didn't include a few clarifying details. Here is the full problem:

I have two rectangles in the plane and I want to find the minimum gap between them. I have a way (read: function) to get the minimum Euclidean distance between them, but the minimum gap is smaller than that (the gap between $(0,0)$ and $(0,2)$ is 1, the point $(0,1)$).

The $x$ and $y$ mentioned in my original problem were the difference vectors between the closest two points of the two rectangles, not as I originally said, the two points themselves.

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    $\begingroup$ Note that $\sqrt{x^2+y^2}$ is not the distance between $x$ and $y$. It is the distance between $(x,y)$ and $(0,0)$. $\endgroup$ – Arnaud Mortier May 30 '18 at 0:10
  • $\begingroup$ There is nothing smaller about $x-1$. If the $x$ component is negative, the vector $(x,y)$ just got larger. $\endgroup$ – Dan Sp. May 30 '18 at 0:23
  • $\begingroup$ @ArnaudMortier Thanks for pointing this out. I edited the post to be more clear about what I was intending. $\endgroup$ – ThePerfectionist May 30 '18 at 0:55
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You can't. $\sqrt{x^2 + y^2}$ is the distance from $(x,y)$ to the point $(0,0)$. $\sqrt{(x-1)^2 + (y-1)^2}$ is the distance to the point $(1,1)$. But there infinitely many points with the same distance from $(0,0)$ (that is, lying on a circle with center $(0,0)$ and a fixed radius) that have different distances to the point $(1,1)$

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Your question is "given that $(x,y)$ lies on some circle centered at the origin, how do I find the distance between $(x,y)$ and $(1,1)$?"

Answer: you don't. The only point in the plane that is equidistant from every point on a circle is the centre of the circle.

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  • $\begingroup$ Though you can put limits on it. If $z\ge 1$ then $z-1\le z' \le z+1$ and if $0 \le z \le 1$ then $1-z \le z' \le 1+z$ $\endgroup$ – Henry May 30 '18 at 0:11
  • $\begingroup$ Sure, the circle is compact, so the distance between $(1,1)$ and a point on it is going to be bounded. But you can't say much more than that. $\endgroup$ – Arnaud Mortier May 30 '18 at 0:13

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