因此，我尝试绘制一个三角形外圆，但我不需要圆心点以外的额外圆。如何绘制部分圆？有没有办法切断这些部分？
这是我目前的绘图代码
\documentclass{article}
\usepackage{tikz}
\usetikzlibrary{through,calc,intersections}
\usepackage{xparse}
\NewDocumentCommand{\bissectrice}{%
O{} % drawing options
mmm % bissector of mmm
m % intersection point between base and bissector
O{1}O{1}% extended drawing of the bissector
}{%
\path[name path=Bis#2#3#4] let
\p1 = ($(#2) - (#3)$),
\p2 = ($(#4) - (#3)$),
\n1 = {veclen(\x1,\y1)/2} ,
\n2 = {veclen(\x2,\y2)/2} ,
\n3 = {max(\n1,\n2)},
\p1 = ($(#3)!\n3!(#2)$),
\p2 = ($(#3)!\n3!(#4)$),
\p3 = ($(\p1) + (\p2) - (#3)$)
in
(#3) -- (\p3) ;
\path[name path = foo] (#2)--(#4) ;
\path[name intersections={of=foo and Bis#2#3#4, by=#5}] ;
\path[#1] ($(#3)!#6!(#5)$) -- ($(#5)!#7!(#3)$) ;
}
\begin{document}
\begin{tikzpicture}[scale=0.3]
\coordinate (A) at (0,0);
\coordinate (A2) at (-3,0);
\coordinate (A1) at (-4,-5);
\coordinate (B) at (4,5);
\coordinate (B1) at (8,10);
\coordinate (B2) at (1,10);
\coordinate (C) at (7,0);
\coordinate (C1) at (10,0);
\coordinate (C2) at (10, -5);
\draw (A)--(B)--(C)--(A);
\draw (A1)--(B1) (B2)--(C2) (C1)--(A2);
\bissectrice {B}{A}{C}{A5};
\bissectrice {A}{B}{C}{B5};
\bissectrice {A}{C}{B}{C5};
\bissectrice {A1}{A}{C}{A3};
\bissectrice {A2}{A}{B}{A4};
\bissectrice {A}{C}{C2}{C3};
\bissectrice {C1}{C}{B}{C4};
\bissectrice {B1}{B}{C}{B3};
\bissectrice {B2}{B}{A}{B4};
\coordinate[label=0:$O_a$] (Oa) at (intersection 1 of A--A3 and C--C3);
\node [draw=black] at (Oa) [circle through={($(A)!(Oa)!(C)$)}]{};
\coordinate[label=-135:$O_b$] (Ob) at (intersection 1 of A--A4 and B4--B);
\node [draw=black] at (Ob) [circle through={($(A)!(Ob)!(B)$)}]{};
\coordinate[label=45:$O_c$] (Oc) at (intersection 1 of B3--B and C--C4);
\node [draw=black] at (Oc) [circle through={($(B)!(Oc)!(C)$)}]{};
\draw (Oa)--node[label=0:$r_a$]{}($(A)!(Oa)!(C)$) (Ob)--node[label=45:$r_b$]{}($(A)!(Ob)!(B)$) (Oc)--node[label=90:$r_c$]{}($(B)!(Oc)!(C)$);
\coordinate (O) at (intersection 1 of A5--A and B5--B);
\draw (O)-- node[label=right:$r$]{}($(A)!(O)!(C)$);
\node [draw=black] at (O) [circle through={($(A)!(O)!(C)$)}]{};
\end{tikzpicture}
\end{document}
我有的图纸：
我想要的图纸：
答案1
正如 Torbjørn 所建议的，你可以绘制clip整个图表。我用一个矩形作为底面，A1并以 和为基础绘制了整个图表B1：
\clip ([shift={(-2,-2)}]A1) rectangle ([shift={(3,-2)}]B1);
但您可以选择另一个剪切路径。
\documentclass{article}
\usepackage{tikz}
\usetikzlibrary{through,calc,intersections}
\usepackage{xparse}
\NewDocumentCommand{\bissectrice}{%
O{} % drawing options
mmm % bissector of mmm
m % intersection point between base and bissector
O{1}O{1}% extended drawing of the bissector
}{%
\path[name path=Bis#2#3#4] let
\p1 = ($(#2) - (#3)$),
\p2 = ($(#4) - (#3)$),
\n1 = {veclen(\x1,\y1)/2} ,
\n2 = {veclen(\x2,\y2)/2} ,
\n3 = {max(\n1,\n2)},
\p1 = ($(#3)!\n3!(#2)$),
\p2 = ($(#3)!\n3!(#4)$),
\p3 = ($(\p1) + (\p2) - (#3)$)
in
(#3) -- (\p3) ;
\path[name path = foo] (#2)--(#4) ;
\path[name intersections={of=foo and Bis#2#3#4, by=#5}] ;
\path[#1] ($(#3)!#6!(#5)$) -- ($(#5)!#7!(#3)$) ;
}
\begin{document}
\begin{tikzpicture}[scale=0.3]
\coordinate (A) at (0,0);
\coordinate (A2) at (-3,0);
\coordinate (A1) at (-4,-5);
\coordinate (B) at (4,5);
\coordinate (B1) at (8,10);
\coordinate (B2) at (1,10);
\coordinate (C) at (7,0);
\coordinate (C1) at (10,0);
\coordinate (C2) at (10, -5);
\clip ([shift={(-2,-2)}]A1) rectangle ([shift={(3,-2)}]B1);
\draw (A)--(B)--(C)--(A);
\draw (A1)--(B1) (B2)--(C2) (C1)--(A2);
\bissectrice {B}{A}{C}{A5};
\bissectrice {A}{B}{C}{B5};
\bissectrice {A}{C}{B}{C5};
\bissectrice {A1}{A}{C}{A3};
\bissectrice {A2}{A}{B}{A4};
\bissectrice {A}{C}{C2}{C3};
\bissectrice {C1}{C}{B}{C4};
\bissectrice {B1}{B}{C}{B3};
\bissectrice {B2}{B}{A}{B4};
\coordinate[label=0:$O_a$] (Oa) at (intersection 1 of A--A3 and C--C3);
\node [draw=black] at (Oa) [circle through={($(A)!(Oa)!(C)$)}]{};
\coordinate[label=-135:$O_b$] (Ob) at (intersection 1 of A--A4 and B4--B);
\node [draw=black] at (Ob) [circle through={($(A)!(Ob)!(B)$)}]{};
\coordinate[label=45:$O_c$] (Oc) at (intersection 1 of B3--B and C--C4);
\node [draw=black] at (Oc) [circle through={($(B)!(Oc)!(C)$)}]{};
\draw (Oa)--node[label=0:$r_a$]{}($(A)!(Oa)!(C)$) (Ob)--node[label=45:$r_b$]{}($(A)!(Ob)!(B)$) (Oc)--node[label=90:$r_c$]{}($(B)!(Oc)!(C)$);
\coordinate (O) at (intersection 1 of A5--A and B5--B);
\draw (O)-- node[label=right:$r$]{}($(A)!(O)!(C)$);
\node [draw=black] at (O) [circle through={($(A)!(O)!(C)$)}]{};
\end{tikzpicture}
\end{document}
