Find the nearest point of intersections of circles
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I need to find the nearest point of intersections of 4 circles. Centers of circles don't change, only the radius of circles changes. I know every intersection of each circle but I don't know how to find the average point of intersections 4 circles. Cannot find intersection of 4 circles, because the measurement error is too high, so I need to find some average point.
![](https://www.mathworks.com/matlabcentral/answers/uploaded_files/1046400/image.png)
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Jan
2022 年 6 月 26 日
How is "average point of intersections 4 circles" mathematically defined? The 4 circles have 12 intersection points. If you have their positions, Matt J's auggestion sounds obvious.
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MJFcoNaN
2022 年 6 月 27 日
Hello,
You may need a different method than intersection. For example create a distance function under your own criterion. This is a simple one:
ori_x = [0 , 0 , -6.19, -6.1];
ori_y = [0 , 4.6 , 4.6 , 0 ];
r = [2.55, 4.05, 6.55, 3.65];
[xx, yy] = ndgrid(-20:1e-1:20,-20:1e-1:20);
vv = NaN([size(xx),4]);
for ii=1:4
vv(:,:,ii) = abs((xx-ori_x(ii)).^2+(yy-ori_y(ii)).^2-r(ii).^2);
end
vs = sum(vv,3);
vs(vs>100) = NaN;
contourf(xx,yy,vs)
[mm,I] = min(vs,[],"all","omitnan");
xx(I)
yy(I)
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その他の回答 (1 件)
Sam Chak
2022 年 6 月 27 日
編集済み: Sam Chak
2022 年 6 月 27 日
Some very simple calculations and math concepts (no looping) that you can definitely follow to find the intersections between the circles. Let's try an example with the red circle and the blue circle.
syms x y
% since red & blue circle aligned at the same y-axis, then pick x as variable
% intersections lie at the top of red circle (+) and bottom of blue circle (–)
redC = sqrt(2.55^2 - (x - 0)^2) + 0;
bluC = - sqrt(4.05^2 - (x - 0)^2) + 4.6;
eqn1 = redC == bluC;
solx = solve(eqn1);
solx = double(solx)
soly = subs(redC, x, solx);
soly = double(soly)
You should be able to find 6 intersections using this concept of coaxial centers.
For non-coaxial centers, let's try finding the intersections between the red circle and the green circle:
% One intersection lie at the top of red circle (+) and bottom of green circle (–)
redC = sqrt(2.55^2 - (x - 0)^2) + 0;
grnC = - sqrt(6.55^2 - (x + 6.19)^2) + 4.6;
eqn2 = redC == grnC;
solx = solve(eqn2);
solx = double(solx)
soly = subs(redC, x, solx);
soly = double(soly)
% One intersection lie at the bottom of red circle (–) and bottom of green circle (–)
redC = - sqrt(2.55^2 - (x - 0)^2) + 0;
grnC = - sqrt(6.55^2 - (x + 6.19)^2) + 4.6;
eqn3 = redC == grnC;
solx = solve(eqn3);
solx = double(solx)
soly = subs(redC, x, solx);
soly = double(soly)
You can find another 6 intersections using this concept of non-coaxial centers.
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