Qfyilyi

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

1. First, we define the circle with the macro tkzDefCircle.


2. This macro returns two values that are the center recovered with the macro tkzGetPoint{O}


3. 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}

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) {};

    }

ifnumX=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

ifnumX=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

ifnumX=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

ifnumX=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_2cdot x_3) + x_3cdot x_4 + y_3cdot (y_4-y_2 )}{x_3cdot y_1 - x_1cdot y_3};.

 ]

 This is texttt{textbackslash n1} in the Tiemph{k}Z style texttt{circle through 3 points}.

 };

fi

ifnumX=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}

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).

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}

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}

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}