Krdytkyu

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}

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}

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 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(pj) {}

    }

    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}

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}