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The search algorithm for the arc, its center, radius and angles is set to 3 points

Given 3 points A, B and C

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How to find the arc starting with A ends in C and passes through B; its central coordinates, radius and angles for r and r '?

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There are several ways to do this. Here is one algorithm:

  • Get your COORDS

    A = {xA, yA}

    B = {xB, yB}

    C = {xC, yC}

    d = {xd, yd}

  • Calculate midpoints of lines AB and BC

    mid_AB = {(xA + xB) / 2, (yA + yB) / 2}

    mid_BC = {(xB + xC) / 2, (yB + yC) / 2}

  • Find the slopes of the lines AB and BC

    slope_AB = (yB-yA) / (xB-xA)

    slope_BC = (yC-yB) / (xC-xB)

  • Build lines through the midpoints of PERPENDICULAR to AB and BC (thanks Yves for the trick!)

    Slope_perp_AB = - (slope_AB) ^ (- 1)

    Slope_perp_BC = - (slope_BC) ^ (- 1)

    *** A line with Slope_perp_AB goes through mid_AB

    *** A line with Slope_perp_BC goes through mid_BC

  • Set the two equations equal to each other and decide to find the intersection! This gives you the point d = {xd, yd} !!!

The calculation of the radius and angles is now trivial with the center point d!

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The center of the circle is equidistant for three given points:

(X-Xa)^2+(Y-Ya)^2 = (X-Xb)^2+(Y-Yb)^2 = (X-Xc)^2+(Y-Yc)^2

Subtracting the first term from the second and third, we get after regrouping:

 2(Xa-Xb) X + 2(Ya-Yb) Y + Xb^2+Yb^2-Xa^2-Ya^2 = 0 2(Xa-Xc) X + 2(Ya-Yc) Y + Xc^2+Yc^2-Xa^2-Ya^2 = 0

This linear system of two equations in two unknowns is easily solved using the Cramer rule.

Radius and angles can be found using the Cartesian-polar transformation around the center:

 R= Sqrt((Xa-X)^2+(Ya-Y)^2) Ta= atan2(Ya-Y, Xa-X) Tc= atan2(Yc-Y, Xc-X)

But you still miss one thing: what is the corresponding part of the arc? Less or more than half the turnover? From Ta to Tb or from Tb to 2 Pi to Ta + 2 Pi , or what? The answer is much less obvious than it seems, try it (because the three angles Ta , Tb and Tc not defined to a multiple of 2 Pi , and you cannot sort them)!

Hint: consider the sign of the region of triangle ABC, exactly half the determinant of the system. It will tell you whether B is on the left or right of the AC.

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Step 1

Find the perpendicular bisector AB and BC.

Step 2

Find the intersection point of these lines.

The point you find will be the center of the circle you want.

Step 3

Calculate the distance of one of the three points from the center found in step 2. This will be the radius of the circle.

NOTE Points A, B, and C must not be on the same line. You must verify this before performing steps 1 to 3.

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The solution to this is almost identical to the solution to the “best fit circle for an undetermined system. Since you have three points that are located exactly on an arc located in a circle centered at (0,0) (given), the system can be solved exactly, and does not require least squares approximation.

 Finding the Center of a Circle Given 3 Points Date: 05/25/2000 at 00:14:35 From: Alison Jaworski Subject: finding the coordinates of the center of a circle Hi, Can you help me? If I have the x and y coordinates of 3 points - ie (x1,y1), (x2,y2) and (x3,y3) - how do I find the coordinates of the center of a circle on whose circumference the points lie? Thank you. Date: 05/25/2000 at 10:45:58 From: Doctor Rob Subject: Re: finding the coordinates of the center of a circle Thanks for writing to Ask Dr. Math, Alison. Let (h,k) be the coordinates of the center of the circle, and r its radius. Then the equation of the circle is: (xh)^2 + (yk)^2 = r^2 Since the three points all lie on the circle, their coordinates will satisfy this equation. That gives you three equations: (x1-h)^2 + (y1-k)^2 = r^2 (x2-h)^2 + (y2-k)^2 = r^2 (x3-h)^2 + (y3-k)^2 = r^2 in the three unknowns h, k, and r. To solve these, subtract the first from the other two. That will eliminate r, h^2, and k^2 from the last two equations, leaving you with two simultaneous linear equations in the two unknowns h and k. Solve these, and you'll have the coordinates (h,k) of the center of the circle. Finally, set: r = sqrt[(x1-h)^2+(y1-k)^2] and you'll have everything you need to know about the circle. This can all be done symbolically, of course, but you'll get some pretty complicated expressions for h and k. The simplest forms of these involve determinants, if you know what they are: |x1^2+y1^2 y1 1| |x1 x1^2+y1^2 1| |x2^2+y2^2 y2 1| |x2 x2^2+y2^2 1| |x3^2+y3^2 y3 1| |x3 x3^2+y3^2 1| h = ------------------, k = ------------------ |x1 y1 1| |x1 y1 1| 2*|x2 y2 1| 2*|x2 y2 1| |x3 y3 1| |x3 y3 1| Example: Suppose a circle passes through the points (4,1), (-3,7), and (5,-2). Then we know that: (h-4)^2 + (k-1)^2 = r^2 (h+3)^2 + (k-7)^2 = r^2 (h-5)^2 + (k+2)^2 = r^2 Subtracting the first from the other two, you get: (h+3)^2 - (h-4)^2 + (k-7)^2 - (k-1)^2 = 0 (h-5)^2 - (h-4)^2 + (k+2)^2 - (k-1)^2 = 0 h^2+6*h+9 - h^2+8*h-16 + k^2-14*k+49 - k^2+2*k-1 = 0 h^2-10*h+25 - h^2+8*h-16 + k^2+4*k+4 - k^2+2*k-1 = 0 14*h - 12*k + 41 = 0 -2*h + 6*k + 12 = 0 10*h + 65 = 0 30*k + 125 = 0 h = -13/2 k = -25/6 Then r = sqrt[(4+13/2)^2 + (1+25/6)^2] = sqrt[4930]/6 Thus the equation of the circle is: (x+13/2)^2 + (y+25/6)^2 = 4930/36 - Doctor Rob, The Math Forum http://mathforum.org/dr.math/

References

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You have three equations to determine the three unknowns xM, yM and R,

 (xA-xM)^2+(yA-yM^2) = R^2

etc .. Subtracting equation A from equations B and C, we obtain

 2*(xB-xA)*xM+2*(yB-yA)*yM = xB^2-xA^2+yB^2-yA^2 2*(xC-xA)*xM+2*(yC-yA)*yM = xC^2-xA^2+yC^2-yA^2

Solving this 2x2-linear system, you get the center point of the circle by inserting a radius into any of the original equation.

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There is a little-known result giving an implicit equation of a circle through 3 points:

 |ZXY 1| |Za Xa Ya 1| |Zb Xb Yb 1| = 0 |Zc Xc Yc 1|

where we defined Z:= X^2 + Y^2 for the sake of brevity.

Calculating the 3x3 minors, we go to:

 M00 Z + M10 X + M20 Y + M30 = 0

and, after normalization, we get the usual equation of the second degree:

 X^2 + Y^2 + 2U X + 2V Y + W = 0

This can be rewritten as:

 (X - U)^2 + (Y - V)^2 = U^2 + V^2 - W

immediately giving the center (U, V) = (-M10/2.M00, -M20/2.M00) and the radius R^2 = U^2 + V^2 - M30/M00 .

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