Let
\begin{eqnarray*}
OA:OB:AC:BC:CD=1:2:\frac{5}{2}:\frac{5}{2}:\frac{5}{2}
\end{eqnarray*}
\begin{cases}
O(0, 0)\\
A(r\cos\theta, r\sin\theta)\\
B(2r, 0)\\
\overrightarrow{AC}=(p,q)^T
\end{cases}
Then
\begin{eqnarray*}
AC^2 = CB^2
\end{eqnarray*}
\begin{eqnarray}
\labeq{condition}
p^2+q^2 &= \left[(r\cos\theta+p)-2r\right]^2 + \left[(r\sin\theta+q)-0\right]^2
\end{eqnarray}
By finding $p$ versus $q$ from eq. (\ref{eq:condition}), we obtain
\begin{equation}
\labeq{pvsq}
p = \frac{(2\sin\theta)q+(5-4\cos\theta)r}{2(2-\cos\theta)}
\end{equation}
Considering the Euclidean norm of $\overrightarrow{AC}$, we have the relational expression between $p$ and $q$.
\begin{equation}
\labeq{norm}
\|\overrightarrow{AC}\|^2=p^2+q^2=\frac{25}{4}r^2
\end{equation}
Inserting eq. (\ref{eq:pvsq}) into eqs. (\ref{eq:norm}) and eliminating $p$, we get
\begin{equation}
\labeq{q}
q = -\frac{r\sin\theta}{2} + r(2-\cos\theta)\sqrt{\frac{5+\cos\theta}{5-4\cos\theta}}
\end{equation}
Substitution of eq. (\ref{eq:q}) in eq. (\ref{eq:pvsq}) leads to eq. (\ref{eq:p}).
\begin{equation}
\labeq{p}
p = \frac{r(2-\cos\theta)}{2}+r\sin\theta\sqrt{\frac{5+\cos\theta}{5-4\cos\theta}}
\end{equation}
On the other hand, the position of joint $C$ and $D$ are
\begin{eqnarray}
\labeq{posC}
\overrightarrow{OC} &=& \overrightarrow{OA} + \overrightarrow{AC} &=& (r\cos\theta+p, r\sin\theta+q)^T\\
\labeq{posD}
\overrightarrow{OD} &=& \overrightarrow{OA} + 2\overrightarrow{AC} &=& (r\cos\theta+2p, r\sin\theta+2q)^T
\end{eqnarray}
Inserting eq. (\ref{eq:q}), (\ref{eq:p}) into eqs. (\ref{eq:posC}), we get
\begin{eqnarray}
\labeq{posC2}
\left\{
\begin{array}{l}
C_x=\frac{r\left(2+\cos\theta\right)}{2}+r\sin\theta\sqrt{\frac{5+\cos\theta}{5-4\cos\theta}} \\
C_y=\frac{r\sin\theta}{2}+r(2-\cos\theta)\sqrt{\frac{5+\cos\theta}{5-4\cos\theta}}
\end{array}
\right.
\end{eqnarray}
Likewise, substitution of eq. (\ref{eq:q}), (\ref{eq:p}) in eqs. (\ref{eq:posD}) leads to eqs. (\ref{eq:posD2}).
\begin{eqnarray}
\labeq{posD2}
\left\{
\begin{array}{l}
D_x=2r+2r\sin\theta\sqrt{\frac{5+\cos\theta}{5-4\cos\theta}} \\
D_y=2r\left(2-\cos\theta\right)\sqrt{\frac{5+\cos\theta}{5-4\cos\theta}}
\end{array}
\right.
\end{eqnarray}
