How to draw a circle (sphere) passing through four points?












3















I am trying to draw a circle (sphere) passing through four points B, C, E, F like this picture enter image description here



I tried with tikz-3dplot and my code



documentclass[border=3mm,12pt]{standalone}
usepackage{fouriernc}
usepackage{tikz,tikz-3dplot}
usepackage{tkz-euclide}
usetkzobj{all}

usetikzlibrary{intersections,calc,backgrounds}

begin{document}

tdplotsetmaincoords{70}{110}
%tdplotsetmaincoords{80}{100}
begin{tikzpicture}[tdplot_main_coords,scale=1.5]
pgfmathsetmacroa{3}
pgfmathsetmacrob{4}
pgfmathsetmacroh{5}

% definitions
path
coordinate(A) at (0,0,0)
coordinate (B) at (a,0,0)
coordinate (C) at (0,b,0)
coordinate (S) at (0,0,h)
coordinate (E) at ({a*h^2/(a*a + h*h)},0,{(a*a*h)/(a*a + h*h)})
coordinate (F) at (0,{(b*h*h)/(b*b + h*h)},{(b*b*h)/(b*b + h*h)});
draw[dashed,thick]
(A) -- (B) (A) -- (C) (A) -- (E) (S)--(A) (F)--(A);
draw[thick]
(S) -- (B) -- (C) -- cycle;
draw[thick]
(F) -- (B) (C)--(E) (F)--(E);
tkzMarkRightAngle(S,E,A);
tkzMarkRightAngle(S,F,A);

foreach point/position in {A/below,B/left,C/below,S/above,E/left,F/above}
{
fill (point) circle (.8pt);
node[position=3pt] at (point) {$point$};
}

end{tikzpicture}
end{document}


and got



enter image description here



How can I draw a circle (sphere) passing through four points B, C, E, F?










share|improve this question


















  • 2





    A circle is already uniquely fixed by 3 (noncollinear) points. There exist answers that show you how to find such a circle. E.g. tex.stackexchange.com/questions/461161/… (Sorry for advertising;-)

    – marmot
    3 hours ago













  • @marmot Is it true in 3D?

    – minhthien_2016
    3 hours ago






  • 1





    I believe that an orthographic projection of a sphere is a circle. The subtle point is whether the projected circle runs through the points you indicate, something that I cannot decide without more information on how the sphere is determined.

    – marmot
    3 hours ago











  • The sphere has centre is midpoint of the segment EC.

    – minhthien_2016
    2 hours ago











  • @marmot The sphere has centre is midpoint of the segment BC, not EC. I am trying your hint.

    – minhthien_2016
    2 hours ago
















3















I am trying to draw a circle (sphere) passing through four points B, C, E, F like this picture enter image description here



I tried with tikz-3dplot and my code



documentclass[border=3mm,12pt]{standalone}
usepackage{fouriernc}
usepackage{tikz,tikz-3dplot}
usepackage{tkz-euclide}
usetkzobj{all}

usetikzlibrary{intersections,calc,backgrounds}

begin{document}

tdplotsetmaincoords{70}{110}
%tdplotsetmaincoords{80}{100}
begin{tikzpicture}[tdplot_main_coords,scale=1.5]
pgfmathsetmacroa{3}
pgfmathsetmacrob{4}
pgfmathsetmacroh{5}

% definitions
path
coordinate(A) at (0,0,0)
coordinate (B) at (a,0,0)
coordinate (C) at (0,b,0)
coordinate (S) at (0,0,h)
coordinate (E) at ({a*h^2/(a*a + h*h)},0,{(a*a*h)/(a*a + h*h)})
coordinate (F) at (0,{(b*h*h)/(b*b + h*h)},{(b*b*h)/(b*b + h*h)});
draw[dashed,thick]
(A) -- (B) (A) -- (C) (A) -- (E) (S)--(A) (F)--(A);
draw[thick]
(S) -- (B) -- (C) -- cycle;
draw[thick]
(F) -- (B) (C)--(E) (F)--(E);
tkzMarkRightAngle(S,E,A);
tkzMarkRightAngle(S,F,A);

foreach point/position in {A/below,B/left,C/below,S/above,E/left,F/above}
{
fill (point) circle (.8pt);
node[position=3pt] at (point) {$point$};
}

end{tikzpicture}
end{document}


and got



enter image description here



How can I draw a circle (sphere) passing through four points B, C, E, F?










share|improve this question


















  • 2





    A circle is already uniquely fixed by 3 (noncollinear) points. There exist answers that show you how to find such a circle. E.g. tex.stackexchange.com/questions/461161/… (Sorry for advertising;-)

    – marmot
    3 hours ago













  • @marmot Is it true in 3D?

    – minhthien_2016
    3 hours ago






  • 1





    I believe that an orthographic projection of a sphere is a circle. The subtle point is whether the projected circle runs through the points you indicate, something that I cannot decide without more information on how the sphere is determined.

    – marmot
    3 hours ago











  • The sphere has centre is midpoint of the segment EC.

    – minhthien_2016
    2 hours ago











  • @marmot The sphere has centre is midpoint of the segment BC, not EC. I am trying your hint.

    – minhthien_2016
    2 hours ago














3












3








3








I am trying to draw a circle (sphere) passing through four points B, C, E, F like this picture enter image description here



I tried with tikz-3dplot and my code



documentclass[border=3mm,12pt]{standalone}
usepackage{fouriernc}
usepackage{tikz,tikz-3dplot}
usepackage{tkz-euclide}
usetkzobj{all}

usetikzlibrary{intersections,calc,backgrounds}

begin{document}

tdplotsetmaincoords{70}{110}
%tdplotsetmaincoords{80}{100}
begin{tikzpicture}[tdplot_main_coords,scale=1.5]
pgfmathsetmacroa{3}
pgfmathsetmacrob{4}
pgfmathsetmacroh{5}

% definitions
path
coordinate(A) at (0,0,0)
coordinate (B) at (a,0,0)
coordinate (C) at (0,b,0)
coordinate (S) at (0,0,h)
coordinate (E) at ({a*h^2/(a*a + h*h)},0,{(a*a*h)/(a*a + h*h)})
coordinate (F) at (0,{(b*h*h)/(b*b + h*h)},{(b*b*h)/(b*b + h*h)});
draw[dashed,thick]
(A) -- (B) (A) -- (C) (A) -- (E) (S)--(A) (F)--(A);
draw[thick]
(S) -- (B) -- (C) -- cycle;
draw[thick]
(F) -- (B) (C)--(E) (F)--(E);
tkzMarkRightAngle(S,E,A);
tkzMarkRightAngle(S,F,A);

foreach point/position in {A/below,B/left,C/below,S/above,E/left,F/above}
{
fill (point) circle (.8pt);
node[position=3pt] at (point) {$point$};
}

end{tikzpicture}
end{document}


and got



enter image description here



How can I draw a circle (sphere) passing through four points B, C, E, F?










share|improve this question














I am trying to draw a circle (sphere) passing through four points B, C, E, F like this picture enter image description here



I tried with tikz-3dplot and my code



documentclass[border=3mm,12pt]{standalone}
usepackage{fouriernc}
usepackage{tikz,tikz-3dplot}
usepackage{tkz-euclide}
usetkzobj{all}

usetikzlibrary{intersections,calc,backgrounds}

begin{document}

tdplotsetmaincoords{70}{110}
%tdplotsetmaincoords{80}{100}
begin{tikzpicture}[tdplot_main_coords,scale=1.5]
pgfmathsetmacroa{3}
pgfmathsetmacrob{4}
pgfmathsetmacroh{5}

% definitions
path
coordinate(A) at (0,0,0)
coordinate (B) at (a,0,0)
coordinate (C) at (0,b,0)
coordinate (S) at (0,0,h)
coordinate (E) at ({a*h^2/(a*a + h*h)},0,{(a*a*h)/(a*a + h*h)})
coordinate (F) at (0,{(b*h*h)/(b*b + h*h)},{(b*b*h)/(b*b + h*h)});
draw[dashed,thick]
(A) -- (B) (A) -- (C) (A) -- (E) (S)--(A) (F)--(A);
draw[thick]
(S) -- (B) -- (C) -- cycle;
draw[thick]
(F) -- (B) (C)--(E) (F)--(E);
tkzMarkRightAngle(S,E,A);
tkzMarkRightAngle(S,F,A);

foreach point/position in {A/below,B/left,C/below,S/above,E/left,F/above}
{
fill (point) circle (.8pt);
node[position=3pt] at (point) {$point$};
}

end{tikzpicture}
end{document}


and got



enter image description here



How can I draw a circle (sphere) passing through four points B, C, E, F?







3d tikz-3dplot






share|improve this question













share|improve this question











share|improve this question




share|improve this question










asked 3 hours ago









minhthien_2016minhthien_2016

1,112815




1,112815








  • 2





    A circle is already uniquely fixed by 3 (noncollinear) points. There exist answers that show you how to find such a circle. E.g. tex.stackexchange.com/questions/461161/… (Sorry for advertising;-)

    – marmot
    3 hours ago













  • @marmot Is it true in 3D?

    – minhthien_2016
    3 hours ago






  • 1





    I believe that an orthographic projection of a sphere is a circle. The subtle point is whether the projected circle runs through the points you indicate, something that I cannot decide without more information on how the sphere is determined.

    – marmot
    3 hours ago











  • The sphere has centre is midpoint of the segment EC.

    – minhthien_2016
    2 hours ago











  • @marmot The sphere has centre is midpoint of the segment BC, not EC. I am trying your hint.

    – minhthien_2016
    2 hours ago














  • 2





    A circle is already uniquely fixed by 3 (noncollinear) points. There exist answers that show you how to find such a circle. E.g. tex.stackexchange.com/questions/461161/… (Sorry for advertising;-)

    – marmot
    3 hours ago













  • @marmot Is it true in 3D?

    – minhthien_2016
    3 hours ago






  • 1





    I believe that an orthographic projection of a sphere is a circle. The subtle point is whether the projected circle runs through the points you indicate, something that I cannot decide without more information on how the sphere is determined.

    – marmot
    3 hours ago











  • The sphere has centre is midpoint of the segment EC.

    – minhthien_2016
    2 hours ago











  • @marmot The sphere has centre is midpoint of the segment BC, not EC. I am trying your hint.

    – minhthien_2016
    2 hours ago








2




2





A circle is already uniquely fixed by 3 (noncollinear) points. There exist answers that show you how to find such a circle. E.g. tex.stackexchange.com/questions/461161/… (Sorry for advertising;-)

– marmot
3 hours ago







A circle is already uniquely fixed by 3 (noncollinear) points. There exist answers that show you how to find such a circle. E.g. tex.stackexchange.com/questions/461161/… (Sorry for advertising;-)

– marmot
3 hours ago















@marmot Is it true in 3D?

– minhthien_2016
3 hours ago





@marmot Is it true in 3D?

– minhthien_2016
3 hours ago




1




1





I believe that an orthographic projection of a sphere is a circle. The subtle point is whether the projected circle runs through the points you indicate, something that I cannot decide without more information on how the sphere is determined.

– marmot
3 hours ago





I believe that an orthographic projection of a sphere is a circle. The subtle point is whether the projected circle runs through the points you indicate, something that I cannot decide without more information on how the sphere is determined.

– marmot
3 hours ago













The sphere has centre is midpoint of the segment EC.

– minhthien_2016
2 hours ago





The sphere has centre is midpoint of the segment EC.

– minhthien_2016
2 hours ago













@marmot The sphere has centre is midpoint of the segment BC, not EC. I am trying your hint.

– minhthien_2016
2 hours ago





@marmot The sphere has centre is midpoint of the segment BC, not EC. I am trying your hint.

– minhthien_2016
2 hours ago










1 Answer
1






active

oldest

votes


















4














A circle is determined by 3 points. A sphere, of course, needs at least 4 points on its boundary to be determined. However, the projection of the sphere, i.e. the circle, won't necessarily run through the projections of these points.



This shows two ways to construct circles that run through some of the points:




  1. The dotted circle runs through F, E and C. It is fixed by this requirement. As a consequence it misses B by a small amount.

  2. The red dashed circle runs through the midpoint of BC and through these points. It misses F and E by small amounts.




documentclass[border=3mm,12pt]{standalone}
usepackage{fouriernc}
usepackage{tikz,tikz-3dplot}
usepackage{tkz-euclide}
usetkzobj{all}
usetikzlibrary{calc,through}
tikzset{circle through 3 points/.style n args={3}{%
insert path={let p1=($(#1)!0.5!(#2)$),
p2=($(#1)!0.5!(#3)$),
p3=($(#1)!0.5!(#2)!1!-90:(#2)$),
p4=($(#1)!0.5!(#3)!1!90:(#3)$),
p5=(intersection of p1--p3 and p2--p4)
in },
at={(p5)},
circle through= {(#1)}
}}

usetikzlibrary{intersections,calc,backgrounds}

begin{document}

tdplotsetmaincoords{70}{110}
%tdplotsetmaincoords{80}{100}
begin{tikzpicture}[tdplot_main_coords,scale=1.5]
pgfmathsetmacroa{3}
pgfmathsetmacrob{4}
pgfmathsetmacroh{5}

% definitions
path
coordinate(A) at (0,0,0)
coordinate (B) at (a,0,0)
coordinate (C) at (0,b,0)
coordinate (S) at (0,0,h)
coordinate (E) at ({a*h^2/(a*a + h*h)},0,{(a*a*h)/(a*a + h*h)})
coordinate (F) at (0,{(b*h*h)/(b*b + h*h)},{(b*b*h)/(b*b + h*h)});
draw[dashed,thick]
(A) -- (B) (A) -- (C) (A) -- (E) (S)--(A) (F)--(A);
draw[thick]
(S) -- (B) -- (C) -- cycle;
draw[thick]
(F) -- (B) (C)--(E) (F)--(E);
tkzMarkRightAngle(S,E,A);
tkzMarkRightAngle(S,F,A);

foreach point/position in {A/below,B/left,C/below,S/above,E/left,F/above}
{
fill (point) circle (.8pt);
node[position=3pt] at (point) {$point$};
}
node[circle through 3 points={F}{E}{C},draw=blue,dotted]{};
draw[red,dashed]
let p1=($(B)-(C)$), n1={veclen(x1,y1)/2} in ($(B)!0.5!(C)$) circle (n1);
end{tikzpicture}
end{document}


enter image description here



It will be possible to construct the sphere as well. However, as mentioned its boundary may not run through any of the points.



ADDENDUM: In your setup, the four points do not determine a unique sphere because they all lie in a plane. Using Mathematica I was able to express F as a linear combination



 F = x B + y C + z E


where



x = -((b^2*h^2)/(a^2*(b^2 + h^2))) 
y = h^2/(b^2 + h^2)
z = (b^2*(a^2 + h^2))/(a^2*(b^2 + h^2))


So in this setup it is not possible to draw a unique sphere.






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    1 Answer
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    active

    oldest

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    active

    oldest

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    4














    A circle is determined by 3 points. A sphere, of course, needs at least 4 points on its boundary to be determined. However, the projection of the sphere, i.e. the circle, won't necessarily run through the projections of these points.



    This shows two ways to construct circles that run through some of the points:




    1. The dotted circle runs through F, E and C. It is fixed by this requirement. As a consequence it misses B by a small amount.

    2. The red dashed circle runs through the midpoint of BC and through these points. It misses F and E by small amounts.




    documentclass[border=3mm,12pt]{standalone}
    usepackage{fouriernc}
    usepackage{tikz,tikz-3dplot}
    usepackage{tkz-euclide}
    usetkzobj{all}
    usetikzlibrary{calc,through}
    tikzset{circle through 3 points/.style n args={3}{%
    insert path={let p1=($(#1)!0.5!(#2)$),
    p2=($(#1)!0.5!(#3)$),
    p3=($(#1)!0.5!(#2)!1!-90:(#2)$),
    p4=($(#1)!0.5!(#3)!1!90:(#3)$),
    p5=(intersection of p1--p3 and p2--p4)
    in },
    at={(p5)},
    circle through= {(#1)}
    }}

    usetikzlibrary{intersections,calc,backgrounds}

    begin{document}

    tdplotsetmaincoords{70}{110}
    %tdplotsetmaincoords{80}{100}
    begin{tikzpicture}[tdplot_main_coords,scale=1.5]
    pgfmathsetmacroa{3}
    pgfmathsetmacrob{4}
    pgfmathsetmacroh{5}

    % definitions
    path
    coordinate(A) at (0,0,0)
    coordinate (B) at (a,0,0)
    coordinate (C) at (0,b,0)
    coordinate (S) at (0,0,h)
    coordinate (E) at ({a*h^2/(a*a + h*h)},0,{(a*a*h)/(a*a + h*h)})
    coordinate (F) at (0,{(b*h*h)/(b*b + h*h)},{(b*b*h)/(b*b + h*h)});
    draw[dashed,thick]
    (A) -- (B) (A) -- (C) (A) -- (E) (S)--(A) (F)--(A);
    draw[thick]
    (S) -- (B) -- (C) -- cycle;
    draw[thick]
    (F) -- (B) (C)--(E) (F)--(E);
    tkzMarkRightAngle(S,E,A);
    tkzMarkRightAngle(S,F,A);

    foreach point/position in {A/below,B/left,C/below,S/above,E/left,F/above}
    {
    fill (point) circle (.8pt);
    node[position=3pt] at (point) {$point$};
    }
    node[circle through 3 points={F}{E}{C},draw=blue,dotted]{};
    draw[red,dashed]
    let p1=($(B)-(C)$), n1={veclen(x1,y1)/2} in ($(B)!0.5!(C)$) circle (n1);
    end{tikzpicture}
    end{document}


    enter image description here



    It will be possible to construct the sphere as well. However, as mentioned its boundary may not run through any of the points.



    ADDENDUM: In your setup, the four points do not determine a unique sphere because they all lie in a plane. Using Mathematica I was able to express F as a linear combination



     F = x B + y C + z E


    where



    x = -((b^2*h^2)/(a^2*(b^2 + h^2))) 
    y = h^2/(b^2 + h^2)
    z = (b^2*(a^2 + h^2))/(a^2*(b^2 + h^2))


    So in this setup it is not possible to draw a unique sphere.






    share|improve this answer






























      4














      A circle is determined by 3 points. A sphere, of course, needs at least 4 points on its boundary to be determined. However, the projection of the sphere, i.e. the circle, won't necessarily run through the projections of these points.



      This shows two ways to construct circles that run through some of the points:




      1. The dotted circle runs through F, E and C. It is fixed by this requirement. As a consequence it misses B by a small amount.

      2. The red dashed circle runs through the midpoint of BC and through these points. It misses F and E by small amounts.




      documentclass[border=3mm,12pt]{standalone}
      usepackage{fouriernc}
      usepackage{tikz,tikz-3dplot}
      usepackage{tkz-euclide}
      usetkzobj{all}
      usetikzlibrary{calc,through}
      tikzset{circle through 3 points/.style n args={3}{%
      insert path={let p1=($(#1)!0.5!(#2)$),
      p2=($(#1)!0.5!(#3)$),
      p3=($(#1)!0.5!(#2)!1!-90:(#2)$),
      p4=($(#1)!0.5!(#3)!1!90:(#3)$),
      p5=(intersection of p1--p3 and p2--p4)
      in },
      at={(p5)},
      circle through= {(#1)}
      }}

      usetikzlibrary{intersections,calc,backgrounds}

      begin{document}

      tdplotsetmaincoords{70}{110}
      %tdplotsetmaincoords{80}{100}
      begin{tikzpicture}[tdplot_main_coords,scale=1.5]
      pgfmathsetmacroa{3}
      pgfmathsetmacrob{4}
      pgfmathsetmacroh{5}

      % definitions
      path
      coordinate(A) at (0,0,0)
      coordinate (B) at (a,0,0)
      coordinate (C) at (0,b,0)
      coordinate (S) at (0,0,h)
      coordinate (E) at ({a*h^2/(a*a + h*h)},0,{(a*a*h)/(a*a + h*h)})
      coordinate (F) at (0,{(b*h*h)/(b*b + h*h)},{(b*b*h)/(b*b + h*h)});
      draw[dashed,thick]
      (A) -- (B) (A) -- (C) (A) -- (E) (S)--(A) (F)--(A);
      draw[thick]
      (S) -- (B) -- (C) -- cycle;
      draw[thick]
      (F) -- (B) (C)--(E) (F)--(E);
      tkzMarkRightAngle(S,E,A);
      tkzMarkRightAngle(S,F,A);

      foreach point/position in {A/below,B/left,C/below,S/above,E/left,F/above}
      {
      fill (point) circle (.8pt);
      node[position=3pt] at (point) {$point$};
      }
      node[circle through 3 points={F}{E}{C},draw=blue,dotted]{};
      draw[red,dashed]
      let p1=($(B)-(C)$), n1={veclen(x1,y1)/2} in ($(B)!0.5!(C)$) circle (n1);
      end{tikzpicture}
      end{document}


      enter image description here



      It will be possible to construct the sphere as well. However, as mentioned its boundary may not run through any of the points.



      ADDENDUM: In your setup, the four points do not determine a unique sphere because they all lie in a plane. Using Mathematica I was able to express F as a linear combination



       F = x B + y C + z E


      where



      x = -((b^2*h^2)/(a^2*(b^2 + h^2))) 
      y = h^2/(b^2 + h^2)
      z = (b^2*(a^2 + h^2))/(a^2*(b^2 + h^2))


      So in this setup it is not possible to draw a unique sphere.






      share|improve this answer




























        4












        4








        4







        A circle is determined by 3 points. A sphere, of course, needs at least 4 points on its boundary to be determined. However, the projection of the sphere, i.e. the circle, won't necessarily run through the projections of these points.



        This shows two ways to construct circles that run through some of the points:




        1. The dotted circle runs through F, E and C. It is fixed by this requirement. As a consequence it misses B by a small amount.

        2. The red dashed circle runs through the midpoint of BC and through these points. It misses F and E by small amounts.




        documentclass[border=3mm,12pt]{standalone}
        usepackage{fouriernc}
        usepackage{tikz,tikz-3dplot}
        usepackage{tkz-euclide}
        usetkzobj{all}
        usetikzlibrary{calc,through}
        tikzset{circle through 3 points/.style n args={3}{%
        insert path={let p1=($(#1)!0.5!(#2)$),
        p2=($(#1)!0.5!(#3)$),
        p3=($(#1)!0.5!(#2)!1!-90:(#2)$),
        p4=($(#1)!0.5!(#3)!1!90:(#3)$),
        p5=(intersection of p1--p3 and p2--p4)
        in },
        at={(p5)},
        circle through= {(#1)}
        }}

        usetikzlibrary{intersections,calc,backgrounds}

        begin{document}

        tdplotsetmaincoords{70}{110}
        %tdplotsetmaincoords{80}{100}
        begin{tikzpicture}[tdplot_main_coords,scale=1.5]
        pgfmathsetmacroa{3}
        pgfmathsetmacrob{4}
        pgfmathsetmacroh{5}

        % definitions
        path
        coordinate(A) at (0,0,0)
        coordinate (B) at (a,0,0)
        coordinate (C) at (0,b,0)
        coordinate (S) at (0,0,h)
        coordinate (E) at ({a*h^2/(a*a + h*h)},0,{(a*a*h)/(a*a + h*h)})
        coordinate (F) at (0,{(b*h*h)/(b*b + h*h)},{(b*b*h)/(b*b + h*h)});
        draw[dashed,thick]
        (A) -- (B) (A) -- (C) (A) -- (E) (S)--(A) (F)--(A);
        draw[thick]
        (S) -- (B) -- (C) -- cycle;
        draw[thick]
        (F) -- (B) (C)--(E) (F)--(E);
        tkzMarkRightAngle(S,E,A);
        tkzMarkRightAngle(S,F,A);

        foreach point/position in {A/below,B/left,C/below,S/above,E/left,F/above}
        {
        fill (point) circle (.8pt);
        node[position=3pt] at (point) {$point$};
        }
        node[circle through 3 points={F}{E}{C},draw=blue,dotted]{};
        draw[red,dashed]
        let p1=($(B)-(C)$), n1={veclen(x1,y1)/2} in ($(B)!0.5!(C)$) circle (n1);
        end{tikzpicture}
        end{document}


        enter image description here



        It will be possible to construct the sphere as well. However, as mentioned its boundary may not run through any of the points.



        ADDENDUM: In your setup, the four points do not determine a unique sphere because they all lie in a plane. Using Mathematica I was able to express F as a linear combination



         F = x B + y C + z E


        where



        x = -((b^2*h^2)/(a^2*(b^2 + h^2))) 
        y = h^2/(b^2 + h^2)
        z = (b^2*(a^2 + h^2))/(a^2*(b^2 + h^2))


        So in this setup it is not possible to draw a unique sphere.






        share|improve this answer















        A circle is determined by 3 points. A sphere, of course, needs at least 4 points on its boundary to be determined. However, the projection of the sphere, i.e. the circle, won't necessarily run through the projections of these points.



        This shows two ways to construct circles that run through some of the points:




        1. The dotted circle runs through F, E and C. It is fixed by this requirement. As a consequence it misses B by a small amount.

        2. The red dashed circle runs through the midpoint of BC and through these points. It misses F and E by small amounts.




        documentclass[border=3mm,12pt]{standalone}
        usepackage{fouriernc}
        usepackage{tikz,tikz-3dplot}
        usepackage{tkz-euclide}
        usetkzobj{all}
        usetikzlibrary{calc,through}
        tikzset{circle through 3 points/.style n args={3}{%
        insert path={let p1=($(#1)!0.5!(#2)$),
        p2=($(#1)!0.5!(#3)$),
        p3=($(#1)!0.5!(#2)!1!-90:(#2)$),
        p4=($(#1)!0.5!(#3)!1!90:(#3)$),
        p5=(intersection of p1--p3 and p2--p4)
        in },
        at={(p5)},
        circle through= {(#1)}
        }}

        usetikzlibrary{intersections,calc,backgrounds}

        begin{document}

        tdplotsetmaincoords{70}{110}
        %tdplotsetmaincoords{80}{100}
        begin{tikzpicture}[tdplot_main_coords,scale=1.5]
        pgfmathsetmacroa{3}
        pgfmathsetmacrob{4}
        pgfmathsetmacroh{5}

        % definitions
        path
        coordinate(A) at (0,0,0)
        coordinate (B) at (a,0,0)
        coordinate (C) at (0,b,0)
        coordinate (S) at (0,0,h)
        coordinate (E) at ({a*h^2/(a*a + h*h)},0,{(a*a*h)/(a*a + h*h)})
        coordinate (F) at (0,{(b*h*h)/(b*b + h*h)},{(b*b*h)/(b*b + h*h)});
        draw[dashed,thick]
        (A) -- (B) (A) -- (C) (A) -- (E) (S)--(A) (F)--(A);
        draw[thick]
        (S) -- (B) -- (C) -- cycle;
        draw[thick]
        (F) -- (B) (C)--(E) (F)--(E);
        tkzMarkRightAngle(S,E,A);
        tkzMarkRightAngle(S,F,A);

        foreach point/position in {A/below,B/left,C/below,S/above,E/left,F/above}
        {
        fill (point) circle (.8pt);
        node[position=3pt] at (point) {$point$};
        }
        node[circle through 3 points={F}{E}{C},draw=blue,dotted]{};
        draw[red,dashed]
        let p1=($(B)-(C)$), n1={veclen(x1,y1)/2} in ($(B)!0.5!(C)$) circle (n1);
        end{tikzpicture}
        end{document}


        enter image description here



        It will be possible to construct the sphere as well. However, as mentioned its boundary may not run through any of the points.



        ADDENDUM: In your setup, the four points do not determine a unique sphere because they all lie in a plane. Using Mathematica I was able to express F as a linear combination



         F = x B + y C + z E


        where



        x = -((b^2*h^2)/(a^2*(b^2 + h^2))) 
        y = h^2/(b^2 + h^2)
        z = (b^2*(a^2 + h^2))/(a^2*(b^2 + h^2))


        So in this setup it is not possible to draw a unique sphere.







        share|improve this answer














        share|improve this answer



        share|improve this answer








        edited 36 secs ago

























        answered 2 hours ago









        marmotmarmot

        91.4k4106199




        91.4k4106199






























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