Intersection of a plane and a sphere with 3dtools - TeX

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user 3.14159

You will need the latest version of the `3dtools` library to run the code below. For the time being I added styles that allow one to record and redraw paths, using the `show path construction` decoration from the `decorations.pathreplacing` library, which one has to load explicitly (at least at this point in time). Please note that there is a much more powerful tool in the making: [spath3](https://github.com/loopspace/spath3). "In the making" is not necessarily an adequate decription, it works beautifully, yet it does, to the best of my knowledge, currently not have the user interface that makes it easy to use for users who are not pgf and expl3 experts at the same time.   

Anyway, the poor tricks here are sufficient to record the paths generated by the nice `tikz-3dplot-circleofsphere` package. So all one has to do is to use the package to construct the circle, and then a simple `even odd rule` fill to fill the plane. 

```
\documentclass[tikz,border=3mm]{standalone}
\usepackage{tikz-3dplot-circleofsphere}
\usetikzlibrary{3dtools}% https://github.com/marmotghost/tikz-3dtools
\usetikzlibrary{decorations.pathreplacing}%
\begin{document}
\tdplotsetmaincoords{60}{110}
\begin{tikzpicture}[tdplot_main_coords,line join = round, 
		line cap = round, declare function={R=5;D=7;}]
  \path  (0,0,3) coordinate (T)  %center of circle defined by the intersection
   (0,0,0) coordinate (I); 
  \path [3d coordinate={(myn)=(T)-(I)}];
  \begin{scope}[canvas is xy plane at z=0,shift={(T)}]
   \path (D,D) coordinate (p1) (D,-D) coordinate (p2)
    (-D,-D) coordinate (p3) (-D,D) coordinate (p4);
  \end{scope}
  \pgfmathsetmacro{\myd}{sqrt(TD("(T)-(I)o(T)-(I)"))}
  \pgfmathsetmacro{\myr}{sqrt(R*R-\myd*\myd)}
  \pgfmathsetmacro{\myaxisangles}{axisangles("(myn)")}
  \pgfmathsetmacro{\myalpha}{{\myaxisangles}[0]}
  \pgfmathsetmacro{\mybeta}{{\myaxisangles}[1]}
  \pgfmathsetmacro{\mygamma}{asin(\myd/R)}
  \tdplotCsDrawCircle[tdplotCsFront/.style={store path=fore},
 		tdplotCsBack/.style={draw,dashed}]{R}{\myalpha}{\mybeta}{\mygamma}  
  \begin{scope}[tdplot_screen_coords]		
    \shade[ball color=white,opacity=0.5] (I) circle[radius=R]; 
    \path [stored path/first coordinate of=fore] coordinate (aux-0)
  		[stored path/last coordinate of=fore]  coordinate (aux-1);
    \fill[opacity=0.3,gray,even odd rule] 
 	 (p1) -- (p2) -- (p3) -- (p4) -- cycle
	 let \p0=(aux-0),\p1=(aux-1),\n0={atan2(\y0,\x0)},\n1={atan2(\y1,\x1)}
	 in	[stored path/restore path=fore] 
	 arc[start angle=\n1,end angle=\n0,radius=R]
	 -- cycle;
    \draw[stored path/restore path=fore];	
  \end{scope}	
\end{tikzpicture}
\end{document}
```
![Screen Shot 2020-12-02 at 6.58.58 PM.png](/image?hash=974421f2029ef9f53fb190aefcc5188de1e155576cf3e703d7937b8b602629e5)

Here is an *experimental* code that is supposed to deal with arbitrary `T`.
```
\documentclass[tikz,border=3mm]{standalone}
\usepackage{tikz-3dplot-circleofsphere}
\usetikzlibrary{3dtools}% https://github.com/marmotghost/tikz-3dtools
\usetikzlibrary{decorations.pathreplacing}%
\begin{document}
\tdplotsetmaincoords{60}{110}
\foreach \mytestT in {(1,2,2),(0,0,-3),(-1,-2,-2),(0,0,3)}
{\begin{tikzpicture}[tdplot_main_coords,line join = round, 
		line cap = round, declare function={R=5;D=7;}]
  \path \mytestT\space  coordinate (T)  %center of circle defined by the intersection
   (0,0,0) coordinate (I); 
  \path [3d coordinate={(myn)=(T)-(I)}];
  \pgfmathsetmacro{\myn}{TD("(myn)")}
  \pgfmathsetmacro{\myd}{sqrt(TD("(myn)o(myn)"))}
  \ifdim\myd pt<0.1pt
   \typeout{The distance between the center of the sphere and the center of the
  	 circle is too small. Cannot determine a unique plane.}
  \else	 
   \pgfmathsetmacro{\mynn}{1/\myd}
   \pgfmathsetmacro{\myez}{TD("\mynn*(\myn)")}
   \pgfmathsetmacro{\myex}{normalcandidate(\myez)}
   \pgfmathsetmacro{\myey}{TD("(\myez)x(\myex)")}
   \path (\myex) coordinate (ex) (\myey) coordinate (ey) (\myez) coordinate (ez);
   \begin{scope}[x={(ex)},y={(ey)},z={(ez)},canvas is xy plane at z=0,shift={(T)}]
	\path (D,D) coordinate (p1) (D,-D) coordinate (p2)
     (-D,-D) coordinate (p3) (-D,D) coordinate (p4);
   \end{scope}
   \pgfmathsetmacro{\mysd}{screendepth(\myn)}
   \pgfmathsetmacro{\myr}{sqrt(R*R-\myd*\myd)}
   \pgfmathsetmacro{\myaxisangles}{axisangles("(myn)")}
   \pgfmathsetmacro{\myalpha}{{\myaxisangles}[0]}
   \pgfmathsetmacro{\mybeta}{{\myaxisangles}[1]}
   \pgfmathsetmacro{\mygamma}{sign(\mysd)*asin(\myd/R)}
   \tdplotCsDrawCircle[tdplotCsFront/.style={draw=none},
 		 tdplotCsBack/.style={draw=none}]{R}{\myalpha}{\mybeta}{\mygamma}		   
   \pgfmathsetmacro{\mydtest}{veclen(x2d("coffs")-x2d("T"),y2d("coffs")-y2d("T"))}
   \ifdim\mydtest pt>2pt
    \pgfmathsetmacro{\mygamma}{-1*\mygamma}
   \fi
   \tikzset{stored path/reset coordinate index}
   \tdplotCsDrawCircle[tdplotCsFront/.style={store path=fore},
 		 tdplotCsBack/.style={draw,dashed}]{R}{\myalpha}{\mybeta}{\mygamma}		   
   \begin{scope}[tdplot_screen_coords]
     % only for animation
     \path (-R-D,-R-D) rectangle (R+D,R+D);	
     \tikzset{draw sphere/.code={\shade[ball color=white,opacity=0.8] (I) circle[radius=R];},
	 draw rest/.code={
	  \ifnum\np>0
	   \path [stored path/first coordinate of=fore] coordinate (aux-0)
  		   [stored path/last coordinate of=fore]  coordinate (aux-1);
	   \begin{scope} 
	    \clip (p1) -- (p2) -- (p3) -- (p4) -- cycle;
    	\fill[opacity=0.3,blue,even odd rule] 
 		 (p1) -- (p2) -- (p3) -- (p4) -- cycle
		 let \p0=(aux-0),\p1=(aux-1),\n0={atan2(\y0,\x0)},
			 \n1={atan2(\y1,\x1)+(\iorder>0?0:360)}
		 in	[stored path/restore path=fore] 
		 arc[start angle=\n1,end angle=\n0,radius=R]
		 -- cycle;
	   \end{scope}	
       \draw[stored path/restore path=fore];
	  \else
       \fill[opacity=0.3,blue,even odd rule] 
 		(p1) -- (p2) -- (p3) -- (p4) -- cycle;
	  \fi 
	 }}
	 \pgfmathtruncatemacro{\np}{tikztdindexoflastcoordinate}
     \pgfmathtruncatemacro{\iorder}{sign(\mysd)}
	 \ifnum\iorder=1
	  \tikzset{draw sphere,draw rest}      	
	 \else
	  \tikzset{draw rest,draw sphere}      	
	 \fi 
   \end{scope}	
   \fi
\end{tikzpicture}}
\end{document}
```
![ani.gif](/image?hash=8faa1144111e29bab566f1f941d94909fb4d9099a85928678346ce55b08e209b)

But really, the other code just gives you some numerical values and does not tell you how they are obtained, at least I do not see that. If it had given you more digits or told you how it is computed, there would not be a problem. There is also a reason why `3dtools` uses internally `fpu`, which is not perfect, but still better than rounding to two digits.

You get the incorrect result because `(O_4)` is incorrect. It should be at `(0.93881,0.66382,1.62599)`, which *rounds* to `(0.94,{2/3},1.63)`, but since the radius of the circle gets computed from the norm of `(O_4)` the additional digits are important. ``` \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,c/.style={circle,fill,inner sep=1pt}, declare function={rs=2;rc=1;d=sqrt(rs*rs-rc*rc);a=sqrt(2); % alpha is just the latitude angle of O_2 and O_3 alpha=asin(rc/rs); % the beta angle emerges from the requirement that O--O_2 and % O--O_3 enclose an angle of 2*alpha beta=acos((cos(2*alpha)-pow(sin(alpha),2))/(pow(cos(alpha),2)));}] \path (0,0,0) coordinate (O) (0,0,d) coordinate (O_1) ({d*cos(alpha)},0,{d*sin(alpha)}) coordinate (O_2) ({d*cos(alpha)*cos(beta)},{d*cos(alpha)*sin(beta)},{d*sin(alpha)}) coordinate (O_3); % basis vectors for circle 2 \pgfmathsetmacro{\myexb}{TDunit("(0,0,1)x(O_2)")} \pgfmathsetmacro{\myeyb}{TDunit("(O_2)x(\myexb)")} % basis vectors for circle 3 \pgfmathsetmacro{\myexc}{TDunit("(0,0,1)x(O_3)")} \pgfmathsetmacro{\myeyc}{TDunit("(O_3)x(\myexc)")} \path[3d coordinate/.list={% {(T_1)=(O_1)+[rc*cos(beta/2)]*(1,0,0)+[rc*sin(beta/2)]*(0,1,0)}, {(T_2)=(O_2)+[rc*cos(90-beta/2)]*(\myexb)+[rc*sin(90-beta/2)]*(\myeyb)}, {(T_3)=(O_3)+[rc*cos(90+beta/2)]*(\myexc)+[rc*sin(90+beta/2)]*(\myeyc)}}] (O)pic[draw=none]{3d circle through 3 points={A={(T_1)},B={(T_2)},C={(T_3)},center name=M}}; \draw[3d/screen coords] (O) circle[radius=rs]; \pgfmathsetmacro{\myM}{TD("(M)")} \pgfmathsetmacro{\myr}{tddistance("(M)","(T_1)")} \typeout{M=(\myM), r=\myr} \path pic{3d/circle on sphere={R=rs,C={(O)}, P={(M)}}}; \path pic{3d/circle on sphere={R=rs,C={(O)}, P={(0,0,d)}}}; \path pic{3d/circle on sphere={R=rs,P={(O_2)}}}; \path pic{3d/circle on sphere={R=rs, P={(O_3)}}}; \path pic{3d/circle on sphere={R=rs, P={(0,0,0)}}}; % % \path foreach \p/\g in {O/-90,O_1/90,O_2/0,O_3/0,T_1/90,T_2/-135,T_3/-45}{ % (\p)node[c]{}+(\g:2.5mm) node{$\p$}}; \end{tikzpicture} \end{document} ```

This is code of the link ``` \documentclass[tikz,border=5mm,convert={outfile=\jobname.png}]{standalone} \usepackage{tikz-3dplot-circleofsphere} \usetikzlibrary{calc,angles} %============== \everymath{\displaystyle} \begin{document} %%%%%%%%%%%%%%%%%%%%%%% \tdplotsetmaincoords{70}{110} \begin{tikzpicture}[tdplot_main_coords,scale=3,font=\footnotesize,>=stealth] \def\r{2} \pgfmathsetmacro{\a}{sqrt(3)} \pgfmathsetmacro{\b}{sqrt(2)} \begin{scope}[thin,black!50] \draw[dashed] (-1.3*\r,0,0) -- (\r,0,0) (0,-1.05*\r,0) -- (0,\r,0) (0,0,0) -- (0,0,\r); \draw[->] (\r,0,0) -- (1.3*\r,0,0) node[anchor=north east] {$x$}; \draw[->] (0,\r,0) -- (0,1.3*\r,0) node[anchor=north] {$y$}; \draw[->] (0,0,\r) -- (0,0,1.3*\r) node[anchor=south east] {$z$}; \draw[tdplot_screen_coords] (\r,0,0) arc (0:180:\r); \tdplotCsDrawLatCircle{\r}{0} \end{scope} \tdplotCsDrawCircle{\r}{0}{60}{60} \tdplotCsDrawCircle{\r}{70.5}{60}{60} \tdplotCsDrawLatCircle{\r}{60} \tdplotCsDrawCircle{\r}{35.4}{35.3}{84.9} \path (0,0,0) coordinate (O) (0,0,\a) coordinate (O_1) (1.5,0,\a/2) coordinate (O_2) (0.5,\b,\a/2) coordinate (O_3) (0.94,{2/3},1.63) coordinate (O_4) (1,0,\a) coordinate (M) (1.09,0.58,1.57) coordinate (K) (barycentric cs:O_1=1,O_2=1,O_3=1) coordinate (L) ($(L)-(O_4)+(K)$) coordinate (N); \draw[dashed] (M)--(O)--(O_1)--(M)--(O_2)--(O_1) (O)--(O_2)--(K)--(O_4)--(O) (O_2)--(L) (K)--(N); \draw[fill=black] (O) circle (0.5pt) node[below right]{$O$} (O_1) circle (0.5pt) node[above left]{$O_1$} (M) circle (0.5pt) node[above]{ $M$} (O_2) circle (0.5pt) node[left]{$O_2$} (O_4) circle (0.5pt) node[right=0.05cm]{\tiny $O_4$} (K) circle (0.5pt) node[left]{$K$} (N) circle (0.5pt) node[below]{ $N$} (L) circle (0.5pt) node[below right]{ $L$}; \end{tikzpicture} %============= \end{document} ``` I do not know how to obtain `(0.94,{2/3},1.63)` coordinate (O_4)`.

Back to my question, I do not know why the circle with center `O_4` incorrect in this code (I see [here](https://www.overleaf.com/project/60b3af145e71b668aa3580e5) ``` \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,c/.style={circle,fill,inner sep=1pt},declare function={rs=2;rc=1;d=sqrt(rs*rs-rc*rc);a=sqrt(2);}] \path (0,0,0) coordinate (O) (0,0,d) coordinate (O_1) (1.5,0,d/2) coordinate (O_2) (0.5,a,d/2) coordinate (O_3) (0.94,{2/3},1.63) coordinate (O_4); \draw[3d/screen coords] (O) circle[radius=rs]; \path pic{3d/circle on sphere={R=rs,C={(O)}, P={(0,0,d)}}}; \path pic{3d/circle on sphere={R=rs,P={(O_2)}}}; \path pic{3d/circle on sphere={R=rs, P={(O_3)}}}; \path pic{3d/circle on sphere={R=rs, P={(O_4)}}}; \path pic{3d/circle on sphere={R=rs, P={(0,0,0)}}}; %\path foreach \p/\g in {O/-90,O_1/90,O_2/0,O_3/0}{(\p)node[c]{}+(\g:2.5mm) node{$\p$}}; \end{tikzpicture} \end{document} ```

Actually it is easy to make it general. ``` \documentclass[tikz,border=3mm]{standalone} \usetikzlibrary{3dtools}% https://github.com/marmotghost/tikz-3dtools \begin{document} \foreach \rc in {1,1.1,...,3.4,3.3,3.2,...,1.1} {\begin{tikzpicture}[3d/install view={phi=110,theta=70}, line cap=butt,line join=round,c/.style={circle,fill,inner sep=1pt}, declare function={rs=4;rc=\rc;d=sqrt(rs*rs-rc*rc); % alpha is just the latitude angle of O_2 and O_3 alpha=90-2*atan2(rc,d); % the beta angle emerges from the requirement that O--O_2 and % O--O_3 enclose an angle of 2*atan2(rc,d) beta=acos((cos(2*atan2(rc,d))-pow(sin(alpha),2))/(pow(cos(alpha),2))); %gamma=acos(cos(beta)+2*cos(alpha)); }] \path (0,0,0) coordinate (O) (0,0,d) coordinate (O_1) ({d*cos(alpha)},0,{d*sin(alpha)}) coordinate (O_2) ({d*cos(alpha)*cos(beta)},{d*cos(alpha)*sin(beta)},{d*sin(alpha)}) coordinate (O_3) ; % basis vectors for circle 2 \pgfmathsetmacro{\myexb}{TDunit("(0,0,1)x(O_2)")} \pgfmathsetmacro{\myeyb}{TDunit("(O_2)x(\myexb)")} % basis vectors for circle 3 \pgfmathsetmacro{\myexc}{TDunit("(0,0,1)x(O_3)")} \pgfmathsetmacro{\myeyc}{TDunit("(O_3)x(\myexc)")} \path[3d coordinate/.list={% {(T_1)=(O_1)+[rc*cos(beta/2)]*(1,0,0)+[rc*sin(beta/2)]*(0,1,0)}, {(T_2)=(O_2)+[rc*cos(90-beta/2)]*(\myexb)+[rc*sin(90-beta/2)]*(\myeyb)}, {(T_3)=(O_3)+[rc*cos(90+beta/2)]*(\myexc)+[rc*sin(90+beta/2)]*(\myeyc)}}] (O)pic{3d circle through 3 points={A={(T_1)},B={(T_2)},C={(T_3)},center name=M}}; \draw[3d/screen coords] (O) circle[radius=rs]; \path pic{3d/circle on sphere={R=rs,C={(O)}, P={(O_1)}}}; \path pic{3d/circle on sphere={R=rs,P={(O_2)}}}; \path pic{3d/circle on sphere={R=rs, P={(O_3)}}}; % \path foreach \p/\g in {O/-90,O_1/90,O_2/0,O_3/0,T_1/90,T_2/-135,T_3/-45}{ % (\p)node[c]{}+(\g:2.5mm) node{$\p$}}; \end{tikzpicture}} \end{document} ``` ![ani.gif](/image?hash=1d30446b3e5068a6500a89a3684968f2ca4d380b7d4f7caf57341c3a69af3b5f)

No, obviously not. But it was not clear to me from the question that it should. Note also that you cannot take `rc/rs` to become too large, at a given point the circles will no longer fit on the circle. I do not have a simple solution for the general case.

You can use the symmetry of the problem, Then you only need to compute two angles, called `alpha` and `beta` in the following code. ``` \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,c/.style={circle,fill,inner sep=1pt}, declare function={rs=2;rc=1;d=sqrt(rs*rs-rc*rc);a=sqrt(2); % alpha is just the latitude angle of O_2 and O_3 alpha=asin(rc/rs); % the beta angle emerges from the requirement that O--O_2 and % O--O_3 enclose an angle of 2*alpha beta=acos((cos(2*alpha)-pow(sin(alpha),2))/(pow(cos(alpha),2)));}] \path (0,0,0) coordinate (O) (0,0,d) coordinate (O_1) ({d*cos(alpha)},0,{d*sin(alpha)}) coordinate (O_2) ({d*cos(alpha)*cos(beta)},{d*cos(alpha)*sin(beta)},{d*sin(alpha)}) coordinate (O_3); % basis vectors for circle 2 \pgfmathsetmacro{\myexb}{TDunit("(0,0,1)x(O_2)")} \pgfmathsetmacro{\myeyb}{TDunit("(O_2)x(\myexb)")} % basis vectors for circle 3 \pgfmathsetmacro{\myexc}{TDunit("(0,0,1)x(O_3)")} \pgfmathsetmacro{\myeyc}{TDunit("(O_3)x(\myexc)")} \path[3d coordinate/.list={% {(T_1)=(O_1)+[rc*cos(beta/2)]*(1,0,0)+[rc*sin(beta/2)]*(0,1,0)}, {(T_2)=(O_2)+[rc*cos(90-beta/2)]*(\myexb)+[rc*sin(90-beta/2)]*(\myeyb)}, {(T_3)=(O_3)+[rc*cos(90+beta/2)]*(\myexc)+[rc*sin(90+beta/2)]*(\myeyc)}}] (O)pic{3d circle through 3 points={A={(T_1)},B={(T_2)},C={(T_3)},center name=M}}; \draw[3d/screen coords] (O) circle[radius=rs]; \path pic{3d/circle on sphere={R=rs,C={(O)}, P={(0,0,d)}}}; \path pic{3d/circle on sphere={R=rs,P={(O_2)}}}; \path pic{3d/circle on sphere={R=rs, P={(O_3)}}}; \path pic{3d/circle on sphere={R=rs, P={(0,0,0)}}}; % % \path foreach \p/\g in {O/-90,O_1/90,O_2/0,O_3/0,T_1/90,T_2/-135,T_3/-45}{ % (\p)node[c]{}+(\g:2.5mm) node{$\p$}}; \end{tikzpicture} \end{document} ``` ![Screen Shot 2021-05-30 at 10.08.49 PM.png](/image?hash=96eef09341f55ceb86fddc19c3037bcf569b18a48dc9d00d1367fefa76dd6732)

@marmot, re: [your answer](#a1788), The circle has center `O_4` incorrect. Note that, the circles with centers `O_1, O_2, O_3` has radius `rc = 1`. My questions: 1) Can we find coordinates 'O_4, O_2, O_3` by using `3dtools`? 2) How to correct the circle with center `O_4`? 3) Can we general this problem?

@marmot, re: [your answer](#a1788), I get some datas from [here](https://www.overleaf.com/project/60b3af145e71b668aa3580e5) and tried it with `3dtools`. My code ``` \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,c/.style={circle,fill,inner sep=1pt},declare function={rs=2;rc=1;d=sqrt(rs*rs-rc*rc);a=sqrt(2);}] \path (0,0,0) coordinate (O) (0,0,d) coordinate (O_1) (1.5,0,d/2) coordinate (O_2) (0.5,a,d/2) coordinate (O_3) (0.94,{2/3},1.63) coordinate (O_4); \draw[3d/screen coords] (O) circle[radius=rs]; \path pic{3d/circle on sphere={R=rs,C={(O)}, P={(0,0,d)}}}; \path pic{3d/circle on sphere={R=rs,P={(O_2)}}}; \path pic{3d/circle on sphere={R=rs, P={(O_3)}}}; \path pic{3d/circle on sphere={R=rs, P={(O_4)}}}; \path pic{3d/circle on sphere={R=rs, P={(0,0,0)}}}; %\path foreach \p/\g in {O/-90,O_1/90,O_2/0,O_3/0}{(\p)node[c]{}+(\g:2.5mm) node{$\p$}}; \end{tikzpicture} \end{document} ``` I got ![ScreenHunter 204.png](/image?hash=d27f20e602c1988b0d8af49d6f6500277da570ed9b48ae9f325913c4f1a500fe)

It might actually not be too difficult to implement this. ``` \documentclass{article} \usepackage[margin=1in]{geometry} \usepackage{amsmath} \usepackage{tikz} \usepackage{tikz-3dplot} \usetikzlibrary{angles,3dtools} \DeclareMathOperator{\sd}{sd} \begin{document} We are given a sphere around a center $C$ or radius $R$ and a point $P$ inside the sphere as well as a normal $n$. How can one draw a circle on the sphere, distinguishing between visible and hidden stretches? \paragraph{\boldmath First test: $|P-C|<R$.} The radius of the circle is $r=\sqrt{R^2-|P-C|^2}$. Call the normal on the screen $n_\mathrm{screen}$. \paragraph{\boldmath Case $n\parallel n_\mathrm{screen}$.} If $n$ and $n_\mathrm{screen}$ are linearly dependent, the circle is completely hidden if the screen depth of $P$ is smaller than the screen depth of $C$, $\sd P <\sd$, and visible otherwise. We can choose a coordinate system in which \[ e_z=\frac{n}{|n|}\;,\quad e_x\text{ arbitary but }\perp e_z \quad\text{and}\quad e_y=e_z\times e_x\] to draw the circle. \paragraph{\boldmath Case $n\not\,\parallel n_\mathrm{screen}$.} From now on we assume that $n$ and $n_\mathrm{screen}$ are not linearly dependent. This means that they span a plane, and we can choose $n_\mathrm{screen}$ to point in the $x$-direction of this plane (see figure~\ref{fig:not_parallel}). \begin{figure}[htb] \centering \begin{tikzpicture} \draw (-3,-3) rectangle (4,3); \draw[dashed] (0,3) coordinate (t) -- (0,-3) coordinate (b)node[below]{$\sd C$}; \draw[-stealth] (3.5,-3.3) -- (4,-3.3) node[right] {$\sd$}; \path (2,-2) coordinate (c1) -- (-1,2) coordinate (c2) node[midway,circle,fill,inner sep=1.2pt,label=above:{$P$}](P){}; \draw[thick,blue] (intersection of c1--c2 and t--b) coordinate (i) -- coordinate (v) (c1) ; \draw[thick,gray] (i) -- coordinate (h) (c2) ; \draw ($(c1)!5mm!90:(c2)$) -- coordinate (r1) ($(c1)!2mm!90:(c2)$) ($(c2)!5mm!-90:(c1)$) -- coordinate (r2) ($(c2)!2mm!-90:(c1)$) (r1) -- node[sloped,fill=white]{$r$} (r2); \draw[red,-stealth] (P) -- ($(P)!20mm!90:(c1)$)coordinate (n) node[pos=1.15]{$n$}; \draw[red,dashed] (n) |- (P) node[pos=0.25,right]{$n_\perp$} coordinate[pos=0.5] (aux) node[pos=0.75,above]{$n_\parallel$} pic[draw,solid,pic text={$\alpha$},angle radius=3em,angle eccentricity=0.6]{angle=c1--P--aux}; \end{tikzpicture} \caption{$n\not\,\parallel n_\mathrm{screen}$.} \label{fig:not_parallel} \end{figure} The normal can be decomposed in a part parallel to $n_\mathrm{screen}$, $n_\parallel$, and a perpendicular part, $n_\perp$. The projection of the circle on this plane is a line, and has slope $\alpha=\arctan(-n_\parallel/n_\perp)$. If $r\cos\alpha>|\sd P|$, the circle has both hidden and visible stretches, from between $-r\cos\alpha$ and $\sd P$ it is hidden and between $\sd P$ and $r\cos\alpha$ it is solid. We are going to draw the circle in a coordinate system defined by \[ e_z=\frac{n}{|n|}\;,\quad e_y=e_z\times n_\mathrm{screen} \quad\text{and}\quad e_x=e_y\times e_z\;.\] The circle is then in the $x$-$y$ plane, and the positive $x$-direction is the direction of increasing screen depth (or visibility). If $r\cos\alpha>|\sd P|$, then the stretch between $\beta:=\left|\arccos\left(\frac{-\sd P}{r\cos\alpha}\right)\right|$ and $-\beta$ is visible, between $\beta$ and $360^\circ-\beta$ it is hidden. \end{document} ``` ![Screen Shot 2020-12-04 at 8.34.47 PM.png](/image?hash=b814232a8efee39f7091f4bab4bd58ebe67a5c2b8b4f8d1cbf3c6d6f1bdc312b)

I did answer that question. But I think you want to ask a different question: given a sphere of radius `R` around a center `C`, a point `P` inside the sphere, and a normal vector `n`, how can one draw a circle on the sphere with center `P` and normal `n` distinguishing hidden and visible stretches. I think this can be done with `3dtools`, so if you ask a separate question I may find a way. But I do not want to change the answer here since I do think it answers the original question.

In this code, I have to do this `\pgfmathsetmacro{\mygamma}{-acos(\rAHK/\myR)}` and `\pgfmathsetmacro{\mygamma}{acos(\rBHK/\myR)}` by hand. That is why, I commented to you how can I make it automatically, yesterday. I also ask you with this code.

``` \documentclass[tikz,border=3mm]{standalone} \usepackage{tikz-3dplot-circleofsphere} \usetikzlibrary{3dtools}% https://github.com/marmotghost/tikz-3dtools \makeatletter \tikzset{ reuse path/.code={\pgfsyssoftpath@setcurrentpath{#1}} } \tikzset{even odd clip/.code={\pgfseteorule}, protect/.code={ \clip[overlay,even odd clip,reuse path=#1] (-6383.99999pt,-6383.99999pt) rectangle (6383.99999pt,6383.99999pt); }} \makeatother \tikzset{reverseclip/.style={overlay,insert path={(-12383.99999pt,-12383.99999pt) rectangle (12383.99999pt,12383.99999pt)}}} \begin{document} \tdplotsetmaincoords{65}{70} \begin{tikzpicture}[tdplot_main_coords,scale=1,line cap=butt, line join=round,declare function={a=5;R=1/2*a;h=a; angB =-30;r=a/2;}] \path (0,0,0) coordinate (A) (0,a,0) coordinate (C) ({ R*cos(angB)}, {a/2 + R*sin(angB)},0) coordinate (B) (0,0,h) coordinate (S) %[3d coordinate={(O)=0.5*(A)+0.5*(C)}]; ; \path[3d/line through={(S) and (B) named lSB}]; \path[3d/line through={(S) and (C) named lSC}]; \path[3d/project={(A) on lSB}] coordinate (H); \path[3d/project={(A) on lSC}] coordinate (K); \path[3d/circumsphere center={A={(A)},B={(B)},C={(C)},D={(H)}}] coordinate (O); % in preparation % \tikzset{3d/sphere with center={(O) and radius R named sO}} \pgfmathsetmacro{\myR}{sqrt(TD("(O)-(A)o(O)-(A)"))} ; \path[3d/circumcircle center={A={(A)},B={(B)},C={(C)}}] coordinate (T); \pgfmathsetmacro{\myr}{sqrt(TD("(A)-(T)o(A)-(T)"))} \pgfmathsetmacro{\Angle}{acos(\myr/\myR)} \begin{scope}[shift={(O)}] \draw[tdplot_screen_coords] (O) circle[radius=\myR]; \end{scope} \path[3d/circumcircle center={A={(H)},B={(B)},C={(K)}}] coordinate (X); \begin{scope}[shift={(O)}] \path[overlay] [3d coordinate={(myn)=(B)-(K)x(B)-(H)}, 3d coordinate={(B-O)=(B)-(O)}]; \pgfmathsetmacro{\myaxisangles}{axisangles("(myn)")} \pgfmathsetmacro{\myalpha}{{\myaxisangles}[0]} \pgfmathsetmacro{\mybeta}{{\myaxisangles}[1]} \pgfmathsetmacro{\rBHK}{sqrt(TD("(X)-(B)o(X)-(B)"))} \pgfmathsetmacro{\mygamma}{acos(\rBHK/\myR)} \tdplotCsDrawCircle[tdplotCsFront/.style={thick, purple}]{\myR}{\myalpha}{\mybeta}{\mygamma} \end{scope} \path[3d/circumcircle center={A={(A)},B={(H)},C={(K)}}] coordinate (Y); \begin{scope}[shift={(O)}] \path[overlay] [3d coordinate={(myn)=(A)-(K)x(A)-(H)}, 3d coordinate={(A-O)=(A)-(O)}]; \pgfmathsetmacro{\myaxisangles}{axisangles("(myn)")} \pgfmathsetmacro{\myalpha}{{\myaxisangles}[0]} \pgfmathsetmacro{\mybeta}{{\myaxisangles}[1]} \pgfmathsetmacro{\rAHK}{sqrt(TD("(Y)-(A)o(Y)-(A)"))} \pgfmathsetmacro{\mygamma}{-acos(\rAHK/\myR)} \tdplotCsDrawCircle[tdplotCsFront/.style={thick, blue}]{\myR}{\myalpha}{\mybeta}{\mygamma} \end{scope} %\path pic{3d circle through 3 points={%A={(H)},B={(B)},C={(C)},center name=MM}}; \draw (S) --(A) (S) -- (H) (S)--(K) ; \draw[dashed] (A) -- (B) (A) -- (C) (B) -- (H) (C) -- (K) (B) -- (C) (A) -- (H) (A) -- (K) (H) -- (K); \foreach \v/\position in { B/below,C/right,A/left,S/above,K/above,H/right,O/below,X/below,Y/below} {\draw[draw =black, fill=black] (\v) circle (1.3pt) node [\position=0.2mm] {$\v$}; } \end{tikzpicture} \end{document} ```

I think the only thing that is wrong is that you put the pics that construct `X` and `Y` into scopes that shift them by `O`. (As long as pgf does not record the 3d coordinates it will be very hard to allow users to define coordinates in different frames and to get reasonable results from 3d computations. In section 1.4 of the `3dtools` manual you can find very rudimentary ideas to solve these problems, but I do not think one can succeed without the support by the maintainers of `tikz`. So you need either to convince them to do that, or to talk to Joseph Wright and persuade him to further develop `l3draw`, and to build in 3d support.)

where is wrong about two points `X` and `Y` in this code? ``` \documentclass[tikz,border=3mm]{standalone} \usepackage{tikz-3dplot-circleofsphere} \usetikzlibrary{3dtools}% https://github.com/marmotghost/tikz-3dtools \makeatletter \tikzset{ reuse path/.code={\pgfsyssoftpath@setcurrentpath{#1}} } \tikzset{even odd clip/.code={\pgfseteorule}, protect/.code={ \clip[overlay,even odd clip,reuse path=#1] (-6383.99999pt,-6383.99999pt) rectangle (6383.99999pt,6383.99999pt); }} \makeatother \tikzset{reverseclip/.style={overlay,insert path={(-12383.99999pt,-12383.99999pt) rectangle (12383.99999pt,12383.99999pt)}}} \begin{document} \tdplotsetmaincoords{65}{70} \begin{tikzpicture}[tdplot_main_coords,scale=1,line cap=butt, line join=round,declare function={a=5;R=1/2*a;h=a; angB =-30;r=a/2;}] \path (0,0,0) coordinate (A) (0,a,0) coordinate (C) ({ R*cos(angB)}, {a/2 + R*sin(angB)},0) coordinate (B) (0,0,h) coordinate (S) %[3d coordinate={(O)=0.5*(A)+0.5*(C)}]; ; \path[3d/line through={(S) and (B) named lSB}]; \path[3d/line through={(S) and (C) named lSC}]; \path[3d/project={(A) on lSB}] coordinate (H); \path[3d/project={(A) on lSC}] coordinate (K); \path[3d/circumsphere center={A={(A)},B={(B)},C={(C)},D={(H)}}] coordinate (O); % in preparation % \tikzset{3d/sphere with center={(O) and radius R named sO}} \pgfmathsetmacro{\myR}{sqrt(TD("(O)-(A)o(O)-(A)"))} ; \path[3d/circumcircle center={A={(A)},B={(B)},C={(C)}}] coordinate (T); \pgfmathsetmacro{\myr}{sqrt(TD("(A)-(T)o(A)-(T)"))} \pgfmathsetmacro{\Angle}{acos(\myr/\myR)} \begin{scope}[shift={(O)}] \draw[tdplot_screen_coords] (O) circle[radius=\myR]; \end{scope} \begin{scope}[shift={(O)}] \path[3d/circumcircle center={A={(H)},B={(B)},C={(K)}}] coordinate (X); \path[overlay] [3d coordinate={(myn)=(B)-(K)x(B)-(H)}, 3d coordinate={(B-O)=(B)-(O)}]; \pgfmathsetmacro{\myaxisangles}{axisangles("(myn)")} \pgfmathsetmacro{\myalpha}{{\myaxisangles}[0]} \pgfmathsetmacro{\mybeta}{{\myaxisangles}[1]} \pgfmathsetmacro{\rBHK}{sqrt(TD("(X)-(B)o(X)-(B)"))} \pgfmathsetmacro{\mygamma}{acos(\rBHK/\myR)} \tdplotCsDrawCircle[tdplotCsFront/.style={thick, purple}]{\myR}{\myalpha}{\mybeta}{\mygamma} \end{scope} \begin{scope}[shift={(O)}] \path[3d/circumcircle center={A={(A)},B={(H)},C={(K)}}] coordinate (Y); \path[overlay] [3d coordinate={(myn)=(A)-(K)x(A)-(H)}, 3d coordinate={(A-O)=(A)-(O)}]; \pgfmathsetmacro{\myaxisangles}{axisangles("(myn)")} \pgfmathsetmacro{\myalpha}{{\myaxisangles}[0]} \pgfmathsetmacro{\mybeta}{{\myaxisangles}[1]} \pgfmathsetmacro{\rAHK}{sqrt(TD("(Y)-(A)o(Y)-(A)"))} \pgfmathsetmacro{\mygamma}{-acos(\rAHK/\myR)} \tdplotCsDrawCircle[tdplotCsFront/.style={thick, blue}]{\myR}{\myalpha}{\mybeta}{\mygamma} \end{scope} %\path pic{3d circle through 3 points={%A={(H)},B={(B)},C={(C)},center name=MM}}; \draw (S) --(A) (S) -- (H) (S)--(K) ; \draw[dashed] (A) -- (B) (A) -- (C) (B) -- (H) (C) -- (K) (B) -- (C) (A) -- (H) (A) -- (K) (H) -- (K); \foreach \v/\position in { B/below,C/right,A/left,S/above,K/above,H/right,O/below,X/below,Y/below} {\draw[draw =black, fill=black] (\v) circle (1.3pt) node [\position=0.2mm] {$\v$}; } \end{tikzpicture} \end{document} ```

How do `A`, `B`, and `C` relate to the present question? A circle is entirely defined by three points, and such a pic already exists. (A circle is also entirely defined by a sphere and a point inside the sphere but not coinciding with the center of the sphere. However, to draw this one need the `tikz-3dplot-circleofsphere` package, at least as of now. )

No. There can be sign depending on whether the center of the circle is "above" or "below" the center of the sphere. This sign depends on the conventions here and there. One could, however, transform in an appropriate frame in which this relation is always correct.

1 Answer