0

I have the coordinates of 3 points through which, a circle should pass . Having the coordinates of the points in 3D, how could I have the coordinates of the center of circumscribed circle ? Also: if one of the points has some deviations and causes a circumscribed circle couldn't pass through the 3 points, is there a way to determine the required coordinates of the third point in a way that the circle could be constructed to find the deviation in space ?

enter image description here

FabioSpaghetti
  • 167
  • 2
    Three noncolinear points always define a unique plane and a unique circle within that plane. “Deviations” won’t prevent this. – amd Dec 28 '19 at 00:02
  • See https://en.wikipedia.org/wiki/Circumscribed_circle#Cartesian_coordinates_from_cross-_and_dot-products for one method. – amd Dec 28 '19 at 00:07
  • The image is not to scale and shows a lie, the drawing contradict the stated side lengths, Shown sides are almost all equial, $a\approx b$ and is much less than $2b$. Use a proper tool, draw it accurately and you will see the problem. This kind of a trick is sometimes used in olympiad questions. – g.kov Mar 01 '20 at 12:03
  • @g.kov These are the XY coordinates of the points : A( 0 , 60 ) , B (-60, 0 ), C(0, -60 ) – FabioSpaghetti Mar 01 '20 at 15:33
  • This case is completely different from the one you mentioned before. Points $A=(0,60),B=(-60,0),C=(0,-60)$ indeed make an isosceles triangle (half of the square), but in this case the side lengths are $b=120$, $a=c=60\times\sqrt2\approx 84.85$. And the center of circumscribed circle is calculated properly: $O=(0,0)$, as expected. – g.kov Mar 01 '20 at 16:52
  • @g.kov you are totally right ! I was mixing up the distances from the mid point of the side with 120 length. thank you – FabioSpaghetti Mar 02 '20 at 10:16

2 Answers2

3

This formula for the center $O$ of a circle is suitable for both 2d and 3d:

\begin{align} O&= A\cdot \frac{a^2\,(b^2+c^2-a^2)}{((b+c)^2-a^2)(a^2-(b-c)^2)} \\ &+B\cdot \frac{b^2\,(a^2+c^2-b^2)}{((a+c)^2-b^2)(b^2-(a-c)^2)} \\ &+C\cdot \frac{c^2\,(b^2+a^2-c^2)}{((b+a)^2-c^2)(c^2-(b-a)^2)} , \end{align}

where $A,B,C$ are the coordinates (2d or 3d) of the three given points, and $a,b,c$ are the side length of the corresponding $\triangle ABC$.

As for deviations of the location of one point, I can see one specific unstable configuration when e.g. $B\approx C$. For $B=C$ the center is $O=\tfrac12(B+C)$, but any small deviation of the location of, say, the point $C$ along the line $AB$ pushes $O$ to infinity, so you can check such a condition and react somewhat reasonably.

g.kov
  • 13,581
  • Thank you very much, but if B(xb,yb) , what should I put for B or C in the equation ? xb or yb ? how do I put both ? didn't get – FabioSpaghetti Feb 20 '20 at 08:20
  • 1
    @FabioSpaghetti: Well, in the given formula $O,A,B,C$ represent the points, that is, a pairs $(O_x,O_y)$, $(A_x,A_y)$, $(B_x,B_y)$, $(C_x,C_y)$ in 2d or the triplets $(O_x,O_y,O_z)$, $(A_x,A_y,A_z)$, $(B_x,B_y,B_z)$,$(C_x,C_y,C_z)$, and $a,b,c$ are the corresponding side lengths (real numbers) of $\triangle ABC$. You can look at the formula as a compact way to represent two (in 2d) or three (in 3d) expressions, that is,

    $O_x=A_x\cdot(\cdots)+B_x\cdot(\cdots)+C_x\cdot(\cdots)$,

    $O_y=A_y\cdot(\cdots)+B_y\cdot(\cdots)+C_y\cdot(\cdots)$,

     – g.kov Feb 20 '20 at 09:10
  • thank you very much – FabioSpaghetti Feb 20 '20 at 09:43
  • There is a problem : if a= 2b = 2c , then this formula does not work – FabioSpaghetti Feb 29 '20 at 23:00
  • @FabioSpaghetti: Condition $a=2,b = 2,c$, means that $A,B,C$ are on the same line, and $O=\infty$. – g.kov Mar 01 '20 at 04:20
  • No I mean an Isosceles triangle with the biggest side twice the others – FabioSpaghetti Mar 01 '20 at 09:36
  • @FabioSpaghetti: Yes, you are talking about a so-called degenerate triangle $ABC$ with $|AB|=c$, $|BC|=a=2c$, $|CA|=b=c$, $A\in BC$, $A=\tfrac12,(B+C)$. The area of this degenerate triangle is zero, and circumradius is $\infty$, location of the center of the circumscribed circle is undefined. – g.kov Mar 01 '20 at 10:35
  • no ! it's not possilbe to be zero ! it's an existing triangle, I upload an image of it to the question ! – FabioSpaghetti Mar 01 '20 at 10:43
  • @FabioSpaghetti: Look: $\cos\angle ABC=\frac{a^2-b^2+c^2}{2ac}$, if $b=c$ we have $\cos\angle ABC=\frac a{2b}$, if also $a=2b$ we have $\cos\angle ABC=\frac {2b}{2b}=1$, hence, $\angle ABC=0$ $\triangle ABC$ is degenerate, its area is zero, $\dots$ – g.kov Mar 01 '20 at 11:38
  • @FabioSpaghetti: Or, even simpler, $b+c=2b=a$, triangle is degenerate, since for a valid triangle we must have $b+c>a$. – g.kov Mar 01 '20 at 11:42
  • thank you , but I was slower than you to upload the image of the triangle ! there's something I don't understand – FabioSpaghetti Mar 01 '20 at 11:42
1

If you have the coordinates three points $A, B,$ and $C,$ you can find the perpendicular bisectors of $AB, AC,$ and $BC$ and then you can find the point of concurrency of the three perpendicular bisectors, which will be the center of the unique circle on which $A, B,$ and $C$ fall.

Soham Konar
  • 424
  • 5
  • 17