答案1
欢迎!你只需要找到椭圆切线为零的点。我们把右椭圆的旋转角度称为alpha
,它的“半径”a
和b
。那么这个角度就是atan(b*cot(alpha)/a)
。下面是一个例子R=1;a=2.7;b=1.8;alpha=30;
,其中R
是圆的半径。
\documentclass[tikz,border=3mm]{standalone}
\begin{document}
\begin{tikzpicture}[declare function={R=1;a=2.7;b=1.8;alpha=30;},
nodes={fill=white,font=\sffamily,align=center}]
\pgfmathsetmacro{\myt}{atan(b*cot(alpha)/a)}
\draw (0,R) circle[radius=R*1cm] (0,2*R) node {test 3\\more\\ text};
\draw[rotate=alpha] (\myt:a*1cm and b*1cm)
circle[x radius=a*1cm,y radius=b*1cm]
(\myt:a*2.5cm and b*1cm) node{test 2\\ more text};
\draw[rotate=180-alpha] (-\myt:a*1cm and b*1cm)
circle[x radius=a*1cm,y radius=b*1cm]
(-\myt:a*2.5cm and b*1cm) node{test 1\\ more text};
%
\begin{scope}[xshift={sqrt(2)*a*2cm+1cm}]
\draw (0,R) circle[radius=R*1cm] (0,2*R) node {test 3};
\draw[rotate=alpha] (\myt:a*1cm and b*1cm)
circle[x radius=a*1cm,y radius=b*1cm]
(\myt:a*2.2cm and b*1cm) node{test 2};
\draw[rotate=180-alpha] (-\myt:a*1cm and b*1cm)
circle[x radius=a*1cm,y radius=b*1cm]
(-\myt:a*2.2cm and b*1cm) node{test 1};
\end{scope}
\end{tikzpicture}
\end{document}
需要调整标签中的值2.5
和2.2
乘积,直到您对结果满意为止。a