Drawing a circle through 3 non-collinear points

Question

Does circle through works with 3 points:

\documentclass{standalone}
\usepackage{tikz}
\usetikzlibrary{calc,through}
\begin{document}
    \begin{tikzpicture}
    \coordinate (A) at (1,1);
    \coordinate (B) at (2,2);
    \coordinate (C) at (3,1.5);

    \node[draw,line width=2pt] [circle through={(A)(B)(C)}] {};

    \foreach \i in {A,B,C} {
        \node[circle,minimum size=1pt,fill=red] at(\i) {};
    }
    \end{tikzpicture}
\end{document}

I want to draw a circle through A,B and C.

AndréC · Accepted Answer · 2018-11-21 17:41:07Z

The tkz-euclide package has a macro to do this. The manual is written in French.

First, we define the circle with the macro \tkzDefCircle.
This macro returns two values that are the center recovered with the macro \tkzGetPoint{O}
and the radius that is recovered with the macro \tkzGetLength{rayon}.

Once this is done, we draw the circle with the macro \tkzDrawCircle[R](O,\rayon pt)

\documentclass[tikz,border=5mm]{standalone}
%\usepackage{tikz}
\usepackage{tkz-euclide}
\usetikzlibrary{calc,through}
\begin{document}
    \begin{tikzpicture}
    \coordinate (A) at (1,1);
    \coordinate (B) at (2,2);
    \coordinate (C) at (3,1.5);

%    \node[draw,line width=2pt] [circle through={(A)(B)(C)}] {};

\tkzDefCircle[circum](A,B,C)
\tkzGetPoint{O} \tkzGetLength{rayon}
\tkzDrawCircle[R](O,\rayon pt)

    \foreach \i in {A,B,C} {
        \node[circle,minimum size=1pt,fill=red] at(\i) {};
    }
    \end{tikzpicture}
\end{document}

score 22 · Accepted Answer · 2018-11-21 21:43:12Z

A new node style, based on the derivation below plus the information that one can use intersection of \p1--\p3 and \p2--\p4, which I learned from AndréC's nice answer

\documentclass[tikz,border=5mm]{standalone}
\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)}
}}

\begin{document}
    \begin{tikzpicture}
    \coordinate (A) at (1,1);
    \coordinate (B) at (2,2);
    \coordinate (C) at (3,1.5);
    \node[circle through 3 points={A}{B}{C},draw=blue]{};
    \foreach \i in {A,B,C} {
        \node[circle,minimum size=1pt,fill=red] at(\i) {};
    }
\end{tikzpicture}
\end{document}

Just for fun: an analytic solution based on calc only. (My personal opinion, though, is that this method is more "TikZy", i.e. closer to how the standard TikZ styles work, than the tkz-euclide macros, which are more like pstricks, which I have left behind. However, this is just a personal opinion, and might not be shared by others.)

\documentclass{standalone}
\usepackage{tikz}
\usetikzlibrary{calc}
\tikzset{circle through 3 points/.style n args={3}{%
insert path={let \p1=($(#1)-(#2)$),\p2=($(#1)!0.5!(#2)$),
    \p3=($(#1)-(#3)$),\p4=($(#1)!0.5!(#3)$),\p5=(#1),\n1={(-(\x2*\x3) + \x3*\x4 + \y3*(-\y2 +
    \y4))/(\x3*\y1 - \x1*\y3)},\n2={veclen(\x5-\x2-\n1*\y1,\y5-\y2+\n1*\x1)} in
    ({\x2+\n1*\y1},{\y2-\n1*\x1}) circle (\n2)}
}}
\begin{document}
    \begin{tikzpicture}
    \coordinate (A) at (1,1);
    \coordinate (B) at (2,2);
    \coordinate (C) at (3,1.5);
    \draw[circle through 3 points={A}{B}{C}];
    \foreach \i in {A,B,C} {
        \node[circle,minimum size=1pt,fill=red] at(\i) {};
    }
    \end{tikzpicture}
\end{document}

(Note that \n1 is a fraction, and could in principle not be well defined. If you ever encounter this case, just change the ordering, e.g. do \draw[circle through 3 points={B}{C}{A}]; or something along those lines.)

ADDENDUM: Explanation of the analytic formula.

\documentclass[tikz,border=3.14mm]{standalone}
\usepackage{tikz}
\usetikzlibrary{calc}
\tikzset{circle through 3 points/.style n args={3}{%
insert path={let \p1=($(#1)-(#2)$),\p2=($(#1)!0.5!(#2)$),
    \p3=($(#1)-(#3)$),\p4=($(#1)!0.5!(#3)$),\p5=(#1),\n1={(-(\x2*\x3) + \x3*\x4 + \y3*(-\y2 +
    \y4))/(\x3*\y1 - \x1*\y3)},\n2={veclen(\x5-\x2-\n1*\y1,\y5-\y2+\n1*\x1)} in
    ({\x2+\n1*\y1},{\y2-\n1*\x1}) circle (\n2)}
}}
\begin{document}
\foreach \X in {1,...,5}
{\begin{tikzpicture}[font=\sffamily]
\path[use as bounding box] (-1,-4) rectangle (6,4);
    \coordinate (A) at (1,1);
    \coordinate (B) at (2,2);
    \coordinate (C) at (3,1.5);
    \foreach \i in {A,B,C} {
        \node[circle,minimum size=1pt,fill=red] at(\i) {};
    }
\ifnum\X=1
 \node[anchor=north,text width=7cm] (start) at (2.5,0){Starting point: 3 points.};
 \foreach \Y in {A,B,C}
 {\draw[-latex] (start) to[out=90,in=-90] (\Y) node[above=2pt]{\Y}; }
\fi
\ifnum\X=2
 \coordinate (auxAB) at ($ (A)!.5!(B) $);
 \coordinate (auxBC) at ($ (B)!.5!(C) $);
 \draw (A) -- (B) -- (C);
 \draw ($ (auxAB)!1.2cm!90:(B) $) -- ($ (auxAB)!1.2cm!-90:(B) $) coordinate(aux1);
 \draw ($ (auxBC)!1.2cm!90:(B) $) coordinate(aux2) -- ($ (auxBC)!1.2cm!-90:(B) $);
 \node[anchor=north,text width=7cm] (int) at (2.5,0){The center of the circle is
 where the lines that run through and are orthogonal to the edges intersect.};
 \draw[-latex] (int) to[out=45,in=-90] (aux1);
 \draw[-latex] (int) to[out=135,in=-90] (aux2);
\fi
\ifnum\X=3
 \coordinate[label=below:$P_2$] (auxAB) at ($ (A)!.5!(B) $);
 \coordinate[label=below:$P_4$] (auxBC) at ($ (B)!.5!(C) $);
 \foreach \Y in {auxAB,auxBC}
 {\fill (\Y) circle (1pt);}
 \draw (A) -- (B) -- (C);
 \draw ($ (auxAB)!1.2cm!90:(B) $) -- ($ (auxAB)!1.2cm!-90:(B) $);
 \draw ($ (auxBC)!1.2cm!90:(B) $) -- ($ (auxBC)!1.2cm!-90:(B) $);
 \node[anchor=north,text width=7cm] (int) at (2.5,0){Call the points in the
 middle $P_2$ and $P_4$, and the differences $P_1=A-B$ and $P_3=B-C$. Then the
 orthogonal lines will fulfill
 \[\gamma_1(\alpha)~=~\left(\begin{array}{c}
 x_2+\alpha\,y_1\\ 
 y_2-\alpha\,x_1\\ 
 \end{array}\right) \]
 and
 \[\gamma_2(\beta)~=~\left(\begin{array}{c}
 x_4+\beta\,y_3\\ 
 y_4-\beta\,x_3\\ 
 \end{array}\right)\;. \]
 };
\fi
\ifnum\X=4
 \coordinate[label=below:$P_2$] (auxAB) at ($ (A)!.5!(B) $);
 \coordinate[label=below:$P_4$] (auxBC) at ($ (B)!.5!(C) $);
 \foreach \Y in {auxAB,auxBC}
 {\fill (\Y) circle (1pt);}
 \draw (A) -- (B) -- (C);
 \draw ($ (auxAB)!1.2cm!90:(B) $) -- ($ (auxAB)!1.2cm!-90:(B) $);
 \draw ($ (auxBC)!1.2cm!90:(B) $) -- ($ (auxBC)!1.2cm!-90:(B) $);
 \node[anchor=north,text width=7cm] (int) at (2.5,0){The center of the circle is
 then simply determined by 
 \[\gamma_1(\alpha)~=~\gamma_2(\beta)\;, \]
 which has the solution
 \[
 \alpha~=~\frac{-(x_2\cdot x_3) + x_3\cdot x_4 + y_3\cdot (y_4-y_2 )}{x_3\cdot y_1 - x_1\cdot y_3}\;.
 \]
 This is \texttt{\textbackslash n1} in the Ti\emph{k}Z style \texttt{circle through 3 points}.
 };
\fi
\ifnum\X=5  
\draw[circle through 3 points={A}{B}{C}];
 \node[anchor=north,text width=7cm] (int) at (2.5,-0.1){Once we have the center,
 determining the radius (\texttt{\textbackslash n2}) is trivial, and we can draw
 the circle with a simple \texttt{insert path}.};
\fi
\end{tikzpicture}}
\end{document}

Yes it is there. I meant outside the code. Mathematical expression. — nidhin, Commented Nov 21, 2018 at 19:25
@nidhin Done. There is a simple animation that you can create with convert -delay 800 -loop 0 -density 300 -alpha remove <pdf> <gif>, where pdf is the pdf file that gets created if you compile the lower code, and gif the name of the animated gif you will create. — user121799, Commented Nov 21, 2018 at 19:54
@blackened This method creates a node of circle shape, and the center can be extracted with <node>.center, e.g. \node[circle through 3 points={A}{B}{C},draw=blue] (n){}; \draw[latex-] (n.center) -- ++ (-30:1); — user121799, Commented Dec 31, 2018 at 6:48

beetlej · Accepted Answer · 2018-11-21 22:26:51Z

Just for fun with @AndréC's answer:

\documentclass[tikz,border=5mm]{standalone}
\usetikzlibrary{calc,through}
\begin{document}
    \begin{tikzpicture}
    \coordinate (A) at (1,1);
    \coordinate (B) at (2,2);
    \coordinate (C) at (3,1.5);

    \draw let \p1=($(A)!0.5!(B)$),
    \p2=($(A)!0.5!(C)$),
    \p3=($(\p1)!2!-90:(B)$),
    \p4=($(\p1)!2!90:(B)$),
    \p5=($(\p2)!2!-90:(C)$),
    \p6=($(\p2)!2!90:(C)$),
    \p7=(intersection of \p3--\p4 and \p5--\p6)
    in
    (A) -- (B)
    (A) -- (C)
    (\p3) -- (\p4)
    (\p5) -- (\p6)
    foreach \j in {1,...,7} {
        node[circle,minimum size=2pt,fill=red,inner sep=0,label=\j] at(\p\j) {}
    }
    node[draw,line  width=1pt,circle through= {(A)}] at (\p7)  {};

\foreach \i in {A,B,C} {
    \node[circle,minimum size=5pt,fill=red,inner sep=0,label=\i] at(\i) {};
}
\end{tikzpicture}
\end{document}

hpekristiansen · Accepted Answer · 2018-11-21 17:52:11Z

You can use tkz-euclide like this:

\documentclass{standalone}
\usepackage{tikz}
\usepackage{tkz-euclide}
\usetikzlibrary{calc,through}
\begin{document}
    \begin{tikzpicture}
    \coordinate (A) at (1,1);
    \coordinate (B) at (2,2);
    \coordinate (C) at (3,1.5);

     \node[draw,line width=2pt] [circle through={(A)(B)(C)}] {};

    \foreach \i in {A,B,C} {
        \node[circle,minimum size=1pt,fill=red] at(\i) {};
    }
   \tkzCircumCenter(A,B,C)\tkzGetPoint{O}
   \tkzDrawCircle(O,A)
 \end{tikzpicture}
\end{document}

(Modified from https://tex.stackexchange.com/a/16024/8650)

If you chose to use tkz-euclide, then you should consider to do all of your drawing with it - depending on what it is - you can e.g. define your points with \tkzDefPoint(x,y).

Display Name · Accepted Answer · 2018-11-21 18:00:06Z

8

Just for comparison purpose.

\documentclass[pstricks]{standalone}  
\usepackage{pst-eucl}
\begin{document}
\foreach \i in {1.0,1.2,...,4.0}{
\begin{pspicture}(-5,-5)(5,5)
    \pstTriangle(4;30){A}(4;90){B}(\i;-45){C}
    \pstCircleABC{A}{B}{C}{O}
\end{pspicture}}
\end{document}

answered Nov 21, 2018 at 18:00

Display Name

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Add a comment |

AndréC · Accepted Answer · 2018-11-21 20:57:29Z

8

Just for fun, another solution inspired by the @marmot solution that calculates the intersection of two defined perpendicular bisector with the calc library.

\documentclass[tikz,border=5mm]{standalone}
\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
                 node at (\p5) [draw,line  width=2pt,circle through= {(#1)}]{}}
}}

\begin{document}
    \begin{tikzpicture}
    \coordinate (A) at (1,1);
    \coordinate (B) at (2,2);
    \coordinate (C) at (3,1.5);
    \draw[circle through 3 points={A}{B}{C}];
    \foreach \i in {A,B,C} {
        \node[circle,minimum size=1pt,fill=red] at(\i) {};
    }
    \end{tikzpicture}
\end{document}

answered Nov 21, 2018 at 20:57

AndréC

26.2k3 gold badges38 silver badges100 bronze badges

Nice! I didn't know that one could use \p1 etc. in intersection of \p1--\p3 and \p2--\p4. (I guess it will be more elegant if you manage to eliminate draw,line width=2pt, from the node since \path[circle through 3 points={A}{B}{C}]; also draws the path, which is not what \path usually does.
– user121799
Commented Nov 21, 2018 at 21:11
@marmot It is simple to delete line width=2pt which is a given parameter of the lucky1928 question. On the other hand, it is more difficult to delete draw. To do that, you have to find a way to pass the parameters to the node, I'll get down to work and explore (finally!) the key handlers. I just saw that you succeeded! Congratulations!
– AndréC
Commented Nov 21, 2018 at 21:56
I really could only do that after I learned the intersection of \p1--\p3 and \p2--\p4 thing from you, which is much better than the analytic computation of the intersection point, and presumably also more stable (even though the analytic derivation was fun ;-). Thanks a lot! (My aim was to create a style that behaves the same way as circle through, which your observation made possible, so thanks again!)
– user121799
Commented Nov 21, 2018 at 22:03
@marmot Thank you very much and I have to return the compliment because it is by adapting your solution that I learned how to use insert path and that I am finally motivated to learn key handlers.
– AndréC
Commented Nov 21, 2018 at 22:07
I really like that you acknowledge other posts (and I try to do the same). I wish all users would do that.... ;-)
– user121799
Commented Nov 21, 2018 at 22:11

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nidhin · Accepted Answer · 2018-11-21 19:05:11Z

The code

\node [draw] at (1,1) [circle through={(A)}] {};

draw a circle whose center is at (1,1) and passes through A. In this case the center of the Circumscribed circle has to be calculated before using through.

I just used the Straightedge and compass construction to calculate the center and then drew circle. The basic idea is that all the perpendicular bisectors of the edges of a triangle meet at the same point: the circumcenter.

\documentclass{standalone}
\usepackage{tikz}
\usetikzlibrary{through,intersections}
\begin{document}
    \begin{tikzpicture}
    \coordinate (A) at (1,1);
    \coordinate (B) at (2,2);
    \coordinate (C) at (3,1.5);
    \path[name path=c1] (A) circle[radius=5cm];
    \path[name path=c2] (B) circle[radius=5cm];
    \path[name path=c3] (C) circle[radius=5cm];
    \path[name intersections={of = c1 and c2}];
    \path[name path=o1] (intersection-1)--(intersection-2);
    \path[name intersections={of = c2 and c3}];
    \path[name path=o2] (intersection-1)--(intersection-2);
    \path[name intersections={of = o1 and o2}];
    \node[draw,line width=2pt] at (intersection-1) [circle through={(A)}]{};

        \foreach \i in {A,B,C} {
           \node[circle,minimum size=1pt,fill=red] at(\i) {};
        }
    \end{tikzpicture}
\end{document}

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