# Intersection of a sphere and a plane knowing equations

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] {$\v$};
}

\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?

• I guess the issue is rather basic: you need to specify the plane in which the circle is in. Your plane has a nontrivial normal vector but you draw the circle in the xy plane, which is why it does not match up. – user121799 Mar 25 at 1:36
• The plane has normal vector is (2,-2,1). – minhthien_2016 Mar 25 at 1:38
• Yes, I know. Naively I would think that it is better to switch to a local coordinate system in which the center of the sphere is at (0,0,0) and the normal goes in the z direction, and then just rotate the view. Do your know this nice answer. It will allow you to draw the intersection in such a way that the visible stretch is distinguished from the hidden one. (If you do not like pgfplot, you could also use this answer.) – user121799 Mar 25 at 1:41

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}

• How can I find the points u and v? The way to find. – minhthien_2016 Mar 25 at 4:16
• @minhthien_2016 They need to be orthogonal to n and to each other. – user121799 Mar 25 at 4:18
• And how about the number 4.01? – minhthien_2016 Mar 25 at 4:23
• I can find u = (1, -4, -10) and v = (8, 7, -2). – minhthien_2016 Mar 25 at 4:25
• Can we reduce the line ({1+2+4.01*cos(\t)*1/sqrt(2)+4.01*sin(\t)*1/sqrt(18)}, {-1-2+4.01*cos(\t)*1/sqrt(2)-4.01*sin(\t)*1/sqrt(18)}, {2+1-4*4.01*sin(\t)*1/sqrt(18)}) as I + n + 4.01* cos(t) * u + 4.01*sin(\t)*v? – minhthien_2016 Mar 25 at 4:34