求两个空间圆之间的正交变换

青青子衿

\begin{align*}
\left\{
\begin{split}
x&=\cos(t)\\
y&=\sin(t)\\
z&=0
\end{split}
\right.
\end{align*}
\begin{align*}
\left\{
\begin{split}
x&=-\dfrac{C\cos (t)}{\sqrt{A^2+C^2}}-\dfrac{AB\sin (t)}{\sqrt{\left(A^2+C^2\right) \left(A^2+B^2+C^2\right)}}\\
y&=\sqrt{\frac{A^2+C^2}{A^2+B^2+C^2}}\,\sin (t)\\
z&=\frac{A \cos (t)}{\sqrt{A^2+C^2}}-\frac{B C \sin (t)}{\sqrt{\left(A^2+C^2\right) \left(A^2+B^2+C^2\right)}}
\end{split}
\right.
\end{align*}
它们之间的曲率与挠率都对应相等，怎么求出它们之间的正交变换呢？

\begin{align*}
\color{black}{\cos(\varphi_{2})=\frac{\left(a^2-b^2-c^2\right)\cos(\varphi_{1})}{a^2+b^2+c^2}+\frac{2 a \sqrt{c^2- \left(b^2+c^2\right)\cos^2(\varphi_{1})}}{a^2+b^2+c^2}}
\end{align*}


\begin{align*}
\color{black}{\cos\langle\boldsymbol{\alpha},\boldsymbol{\beta}\rangle=-\frac{\left(a^{2}-b^{2}-c^{2}\right)\left(a^{2}+b^{2}+c^{2}\right)}{4\left(b^{2}+c^{2}\right)}-\frac{a^{2}\left(a^{2}+b^{2}+c^{2}\right)\cos\left(\varphi_{1}\right)\cos\left(\varphi_{2}\right)}{\left(b^{2}+c^{2}\right)\left(\cos\left(\varphi_{1}\right)-\cos\left(\varphi_{2}\right)\right)^{2}} }
\end{align*}

\begin{align*}
\color{black}{
R=\sqrt{\frac{\left(a^{2}+b^{2}+c^{2}\right)^{2}}{4\left(b^{2}+c^{2}\right)}+\frac{a^{2}\left(a^{2}+b^{2}+c^{2}\right)\cos\left(\varphi_{1}\right)\cos\left(\varphi_{2}\right)}{\left(b^{2}+c^{2}\right)\left(\cos\left(\varphi_{1}\right)-\cos\left(\varphi_{2}\right)\right)^{2}} }
}
\end{align*}

\begin{align*}
\color{black}{\cos(T)=\frac{\left(b^{2}+c^{2}-a^{2}\right)\cos\left(\varphi_{1}\right)-\left(a^{2}+b^{2}+c^{2}\right)\cos\left(\varphi_{2}\right)}{2a\cos\left(\varphi_{1}\right)} }
\end{align*}