Intersection of a sphere and a plane knowing equations

Question

I am trying draw a circle is intersection of a plane has equation 2 x − 2 y + z − 15 = 0 and the equation of the sphere is ( x − 1)^2 + ( y + 1)^ 2 + ( z − 2)^ 2 − 25 = 0.

The plane cut the sphere is a circle with centre (3,-3,3 and radius r = 4.

I can't draw the circle. I tried

\documentclass[12pt,border = 2 mm]{standalone}
\usepackage{tikz}
\usepackage{tikz-3dplot}
\usetikzlibrary{arrows,calc,backgrounds}
\begin{document}
\tdplotsetmaincoords{60}{110}
\begin{tikzpicture}[tdplot_main_coords]
\path
      coordinate (T) at (3,-3,3)
      coordinate (I) at (1,-1,2);

\foreach \v/\position in {T/above,I/below} {
    \draw[fill=black] (\v) circle (0.7pt) node [\position=0.2mm] {$\v$};
}

\draw[dashed] (T) circle[radius={4}];

\begin{scope}[tdplot_screen_coords, on background layer]
  \pgfmathsetmacro{\R}{5}%
  \fill[ball color=purple, opacity=1.0] (I) circle (\R); 
\end{scope}
\end{tikzpicture}
\end{document} 

How can I draw the circle?

Answer 1

We know the normal n of the plane and the radius of the circle. Call two vectors that are orthogonal to n and orthogonal to each other u and v. Then the circle is given by

 gamma(t) = I + n + r * cos(t) * u + r* sin(t) * v,

where n fulfills the length constraint n^2+r^2=R^2, where R is the radius of the sphere and r the radius of the circle. Your vector n satisfies this constraint already, so we do not have to change its normalization.

Now we need to figure out what the visible and invisible stretches are. Any point on the sphere has either a positive or negative projection on the normal vector on the screen

n_screen =({sin(\tdplotmaintheta)*sin(\tdplotmainphi)},{-sin(\tdplotmaintheta)*cos(\tdplotmainphi)},{cos(\tdplotmaintheta)});

So we need to find the zeros of the projection gamma(t).n_screen. This can be accomplished by letting TikZ find the intersections. Of course we do not really draw the paths here, and use overlay for them not to screw up the bounding box. Notice that the current version assumes that there are two zeros, so if you drastically change the view angles, this version won't work any more.

\documentclass[12pt,border=2mm,tikz]{standalone}
\usepackage{tikz-3dplot}
\usetikzlibrary{arrows,calc,backgrounds,intersections}
\begin{document}
\tdplotsetmaincoords{60}{110}
\begin{tikzpicture}[tdplot_main_coords,
declare function={dicri(\t,\th,\ph,\R)=
sin(\th)*sin(\ph)*(2+\R*cos(\t)*1/sqrt(2)+\R*sin(\t)*1/sqrt(18))-
sin(\th)*cos(\ph)*(-2+\R*cos(\t)*1/sqrt(2)-\R*sin(\t)*1/sqrt(18))+
cos(\th)*(1-4*\R*sin(\t)*1/sqrt(18));}]
\path
      coordinate (T) at (3,-3,3)
      coordinate (I) at (1,-1,2)
      coordinate (n) at (2,-2,1)
      coordinate (u) at ({1/sqrt(2)},{1/sqrt(2)},0)
      coordinate (v) at ({1/sqrt(18)},{-1/sqrt(18)},{-4/sqrt(18)});
      % the coordinatesn, u and v are not really used here

\foreach \v/\position in {T/above,I/below} {
    \draw[fill=black] (\v) circle (0.7pt) node [\position=0.2mm] {$\v$};
}
% \draw[red,thick,-latex] (0,0,0) --
% ({sin(\tdplotmaintheta)*sin(\tdplotmainphi)},
% {-sin(\tdplotmaintheta)*cos(\tdplotmainphi)},{cos(\tdplotmaintheta)});
% normal to screen
\begin{scope}[tdplot_screen_coords, on background layer]
  \pgfmathsetmacro{\R}{5}%
  \fill[ball color=purple, opacity=1.0] (I) circle (\R);
  % determine the zeros of dicri
  \path[overlay,name path=dicri] plot[variable=\x,domain=0:360,samples=73]
  ({\x*1pt},{dicri(\x,\tdplotmaintheta,\tdplotmainphi,4)});
  \path[overlay,name path=zero] (0,0) -- (360pt,0);
  \path[name intersections={of=dicri and zero,total=\t}]
  let \p1=(intersection-1),\p2=(intersection-2) in
  \pgfextra{\xdef\xmin{\x1}\xdef\xmax{\x2}};
\end{scope}
\pgfmathsetmacro{\R}{4}
\draw[dashed] plot[variable=\t,domain=\xmin:\xmax,samples=73,smooth] 
 ({1+2+\R*cos(\t)*1/sqrt(2)+\R*sin(\t)*1/sqrt(18)},
 {-1-2+\R*cos(\t)*1/sqrt(2)-\R*sin(\t)*1/sqrt(18)},
 {2+1-4*\R*sin(\t)*1/sqrt(18)});
\draw[thick] plot[variable=\t,domain=\xmax:\xmin+360,samples=73,smooth] 
 ({1+2+\R*cos(\t)*1/sqrt(2)+\R*sin(\t)*1/sqrt(18)},
 {-1-2+\R*cos(\t)*1/sqrt(2)-\R*sin(\t)*1/sqrt(18)},
 {2+1-4*\R*sin(\t)*1/sqrt(18)});
\end{tikzpicture}
\end{document}

And here is a plane, using your nicer vectors u and v from the chat.

\documentclass[12pt,border=2mm,tikz]{standalone} 
\usepackage{tikz-3dplot} 
\usetikzlibrary{arrows,calc,backgrounds,intersections} 
\makeatletter % https://tex.stackexchange.com/a/38995/121799
\tikzset{
  use path/.code={\pgfsyssoftpath@setcurrentpath{#1}}
}
\makeatother
\begin{document} 
\tdplotsetmaincoords{60}{110} 
\begin{tikzpicture}[tdplot_main_coords, 
  declare function={dicri(\t,\th,\ph,\R)=% 
  sin(\th)*sin(\ph)*(2+\R*cos(\t)/3+2*\R*sin(\t)/3)-%
  sin(\th)*cos(\ph)*(-2 +2*\R*cos(\t)/3 + \R*sin(\t)/3)+%
  cos(\th)*(1+2*\R*cos(\t)/3-2*\R*sin(\t)/3);}] 
  \pgfmathsetmacro{\R}{5}% 
  \path  coordinate (T) at (3,-3,3) 
   coordinate (I) at (1,-1,2) 
   coordinate (n) at (2,-2,1) 
   coordinate (u) at (1, 2, 2) 
   coordinate (v) at (2, 1, -2); 
  % the coordinatesn, u and v are not really used here 
   \path[tdplot_screen_coords,shift={(I)},use as bounding box] (-1.2*\R,-1.2*\R)rectangle (1.2*\R,1.2*\R);   

  \foreach \v/\position in {T/above,I/below} { 
   \draw[fill=black] (\v) circle (0.7pt) node [\position=0.2mm] {$\v$}; 
  } 
  % \draw[red,thick,-latex] (0,0,0) -- 
  % ({sin(\tdplotmaintheta)*sin(\tdplotmainphi)}, 
  % {-sin(\tdplotmaintheta)*cos(\tdplotmainphi)},{cos(\tdplotmaintheta)}); 
  % normal to screen 
  \begin{scope}[tdplot_screen_coords, on background layer] 
   \fill[ball color=green, opacity=0.8] (I) circle (\R); 
   % determine the zeros of dicri 
   \path[overlay,name path=dicri] plot[variable=\x,domain=0:360,samples=73] 
   ({\x*1pt},{dicri(\x,\tdplotmaintheta,\tdplotmainphi,4)}); 
   \path[overlay,name path=zero] (0,0) -- (360pt,0); 
   \path[name intersections={of=dicri and zero,total=\t}] 
   let \p1=(intersection-1),\p2=(intersection-2) in 
   \pgfextra{\xdef\tmin{\x1}\xdef\tmax{\x2}}; 
  \end{scope} 
  \pgfmathsetmacro{\SmallR}{4} 
  \draw[dashed] plot[variable=\t,domain=\tmin:\tmax,samples=50,smooth] 
   ({1+2+\SmallR*cos(\t)/3+2*\SmallR*sin(\t)/3}, 
   {-1-2 +2*\SmallR*cos(\t)/3+ \SmallR*sin(\t)/3}, 
   {2+1+2*\SmallR*cos(\t)/3 - 2*\SmallR*sin(\t)/3 }); 
  \draw[thick,save path=\pathA] plot[variable=\t,domain=\tmax:\tmin+360,samples=50,smooth] 
   ({1+2+\SmallR*cos(\t)/3+2*\SmallR*sin(\t)/3}, 
   {-1-2 +2*\SmallR*cos(\t)/3+ \SmallR*sin(\t)/3}, 
   {2+1+2*\SmallR*cos(\t)/3 - 2*\SmallR*sin(\t)/3 }); 
  \path ({1+2+\SmallR*cos(\tmin)/3+2*\SmallR*sin(\tmin)/3}, 
   {-1-2 +2*\SmallR*cos(\tmin)/3+ \SmallR*sin(\tmin)/3}, 
   {2+1+2*\SmallR*cos(\tmin)/3 - 2*\SmallR*sin(\tmin)/3 }) coordinate (pmin)
   ({1+2+\SmallR*cos(\tmax)/3+2*\SmallR*sin(\tmax)/3}, 
   {-1-2 +2*\SmallR*cos(\tmax)/3+ \SmallR*sin(\tmax)/3}, 
   {2+1+2*\SmallR*cos(\tmax)/3 - 2*\SmallR*sin(\tmax)/3 }) coordinate (pmax);
  \begin{scope}[tdplot_screen_coords]
   \clip[shift={(I)}] (-1.2*\R,-1.2*\R)rectangle (1.2*\R,1.2*\R);   
   \path[fill=gray,fill opacity=0.4,even odd rule] let \p1=($(pmin)-(I)$),\p2=($(pmax)-(I)$),
   \p3=($(pmax)-(pmin)$),\n1={atan2(\y1,\x1)},\n2={atan2(\y2,\x2)},
   \n3={atan2(\y3,\x3)}
    in [use path=\pathA]  (pmin) arc(\n1:\n2-360:\R) 
    (0,-6) -- ++(\n3:{12cm/sin(\n3)}) -- ++(\n3+90:{12cm/sin(\n3)})
    -- ++(\n3+180:{12cm/sin(\n3)}) -- cycle;
  \end{scope}
\end{tikzpicture} 
\end{document}

Answer 2

Many thanks to Schrödinger's cat about 3dtools. In this answer, I use Mathematica to find coordinates of three points A, B, C and draw circle (ABC). I use tikz-3dplot-circleofsphere at here to draw style of line of the circle (ABC).

\documentclass[12pt,tikz,border=2 mm]{standalone}
\usepackage{tikz-3dplot-circleofsphere}
\usetikzlibrary{3dtools}
\begin{document}
    \tdplotsetmaincoords{60}{110}
\begin{tikzpicture}[scale=1,tdplot_main_coords,declare function={R=5;
}]
    \path   (3,-3,3) coordinate (T)
    (1,-1,2) coordinate (I) 
    (1, {1/5 *(-23 - 2* sqrt(11))}, {1/5 *(19 - 4 *sqrt(11))}) coordinate (A)
    (1, {1/5 *(-23 + 2* sqrt(11))}, {1/5 *(19 + 4 *sqrt(11))}) coordinate (B)
    ({1/4* (13 + sqrt(119))}, {1/4 *(-13 + sqrt(119))}, 2) coordinate (C);
\begin{scope}[tdplot_screen_coords] 
\fill[ball color=green, opacity=0.8] (I) circle (R);
    \end{scope}
\begin{scope}[shift={(I)}]
\path[overlay] [3d coordinate={(A-B)=(A)-(B)},
3d coordinate={(A-C)=(A)-(C)},
3d coordinate={(myn)=(A-B)x(A-C)},
3d coordinate={(A-T)=(A)-(T)}];
\pgfmathsetmacro{\myaxisangles}{axisangles("(myn)")}
\pgfmathsetmacro{\myalpha}{{\myaxisangles}[0]}
\pgfmathsetmacro{\mybeta}{{\myaxisangles}[1]}
\pgfmathsetmacro{\mygamma}{-acos(sqrt(TD("(A-T)o(A-T)"))/R)}
\tdplotCsDrawCircle[tdplotCsFront/.style={thick}]{R}{\myalpha}{\mybeta}{\mygamma} 
\end{scope} 

\foreach \p in {I,T,B}
\draw[fill=black] (\p) circle (1.5 pt);
\foreach \p/\g in {I/0,T/-90,B/30}
\path (\p)+(\g:3mm) node{$\p$};
\draw[dashed] (I) -- (T) -- (B) -- cycle;
    \end{tikzpicture}
\end{document} 

Answer 3

I add this because you asked me to use the 3d circle through 3 points pic here. It can be used after we have three points. These points can be constructed by adding vectors of length 4, the radius, to T which lie in the plane of the circle. These vectors are taken from the other answer, and rescaled.

\documentclass[12pt,tikz,border=2 mm]{standalone}
\usepackage{tikz-3dplot}
\usetikzlibrary{3dtools,backgrounds}
\begin{document}
\tdplotsetmaincoords{60}{110}
\begin{tikzpicture}[tdplot_main_coords]
\pgfmathsetmacro{\mysq}{4/3}
\path   (3,-3,3) coordinate (T)
      (1,-1,2) coordinate (I) 
      (1, 2, 2) coordinate (u)  
      (2, 1, -2) coordinate (v)
      [3d coordinate={(A)=(T)+\mysq*(u)}]
      [3d coordinate={(B)=(T)+\mysq*(v)}]
      [3d coordinate={(C)=(T)-\mysq*(u)}];

\foreach \v/\position in {T/above,I/below} {
    \draw[fill=black] (\v) circle (0.7pt) node [\position=0.2mm] {$\v$};
}

\path[dashed] pic{3d circle through 3 points={A={(A)},B={(B)},C={(C)}}};

\begin{scope}[tdplot_screen_coords, on background layer]
  \pgfmathsetmacro{\R}{5}%
  \fill[ball color=purple, opacity=1.0] (I) circle (\R); 
\end{scope}
\end{tikzpicture}
\end{document} 

Answer 4

As an alternative, Asymptote version:

// spherexplane.asy
//
// run 
//   asy -f pdf -render=4 -noprc spherexplane.asy
// to get a standalone raster spherexplane.pdf
//
import solids;
size(8cm); size3(100,100);
currentprojection=orthographic(camera=(66,40,-9),zoom=0.9);
currentlight=Headlamp;

pen linePen=darkblue+1.3bp;
pen dotPen= darkblue+3bp;
pen dashPen=1bp+linetype(new real[]{4,3})+linecap(0);

// Eqn of the sphere (x - 1)^2 + (y + 1)^ 2 + (z - 2)^ 2 - 25 = 0
triple O=(1,-1,2);
real R=5;

// Eqn of the plane  2 x - 2 y + z - 15 = 0
triple fp(real x, real y){return (x,y,- 2 x + 2 y + 15);}

triple Np=unit((2,-2,1)); // plane normal
triple A=fp(0,0);  // any point on the plane
triple C=O+dot(A-O,Np)*Np; // center of the circle cross section

real d=abs(C-O);
real r=sqrt(R^2-d^2);

guide3 baseArc=Arc(O,O+Np*R,O-Np*R,normal=cross(Z,Np));
revolution b=revolution(O,baseArc,axis=Np); // spherical surface
skeleton s;
real t=acos(d/R)/pi; // fraction of the arc length at the cutting point
b.transverse(s,reltime(b.g,t),P=currentprojection);
guide3 circCut=s.transverse.back[0] & s.transverse.front[0] & cycle;

triple D=relpoint(circCut,0.6);

draw(surface(b),paleblue+opacity(0.3));
draw(surface(circCut),orange+opacity(0.3));

draw(s.transverse.front,linePen);
draw(s.transverse.back, dashPen);

draw(O--C--D, dashPen);

dot("$O$",O,dotPen);
dot(Label("$C$",unit(C-O)),C,dotPen);
dot("$D$",D,dotPen);

xaxis3(xmin=0,xmax=1,red,above=true);
yaxis3(ymin=0,ymax=1,deepgreen,above=true);
zaxis3(zmin=0,zmax=1,blue,above=true);

Answer 5

I use 3dtools like this.

\documentclass[tikz,border=3mm]{standalone}
\usetikzlibrary{3dtools}% https://github.com/marmotghost/tikz-3dtools
\begin{document}
    \begin{tikzpicture}[3d/install view={phi=110,theta=70},line cap=butt,line join=round,declare function={R=5;},c/.style={circle,fill,inner sep=1pt}] 
    \path[overlay]
        (0,0,15) coordinate (A)
        (1,-1,2)  coordinate (I);
        \draw[3d/screen coords] (I) circle[radius=R]; 
        \path[overlay,3d/plane with normal={(2,-2,1) through (A) named p}];
        \path[3d/project={(I) on p}] coordinate (T);
\pic{3d/circle on sphere={R=R,C={(I)}, P={(T)}}};
 \path foreach \p/\g in {I/-90,T/90}
 {(\p)node[c]{}+(\g:2.5mm) node{$\p$}};
\end{tikzpicture}
\end{document}