Let $\Phi_1$ be the anti-clockwise rotation in the plane about $(0,0)$ by an angle $\Theta$ and $\Phi_2$ be the anti-clockwise rotation in the plane about $(2,0)$ by an angle $\Theta$. If $\Theta\neq n\pi$ for any integer $n$, then show that $\Phi_2\circ \Phi_1$ is a rotation about the point $\big(1, -\tan \frac{\Theta}{2}\big).$
My attempt: We can write $$\Phi_1(x,y)=\begin{bmatrix} \cos \Theta& -\sin \Theta\\ \sin\Theta&\cos\Theta\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}.$$ Also,about $(2,0)$, we can write $\Phi_2$ as $$\Phi_2(x,y)=\begin{bmatrix} \cos \Theta& -\sin \Theta\\ \sin\Theta&\cos\Theta\end{bmatrix}\begin{bmatrix}x-2\\y\end{bmatrix}+\begin{bmatrix}2\\0\end{bmatrix}.$$
Now, for composition, we have $$\Phi_2\circ \Phi_1(x)=\Phi_2\bigg(\begin{bmatrix} x\cos \Theta -y\sin \Theta\\ x\sin\Theta+y\cos\Theta\end{bmatrix}\bigg)\ \ \ \ \ \ \ \ $$ $$=\begin{bmatrix} \cos \Theta& -\sin \Theta\\ \sin\Theta&\cos\Theta\end{bmatrix}\begin{bmatrix} x\cos \Theta -y\sin \Theta-2\\ x\sin\Theta+y\cos\Theta\end{bmatrix}+\begin{bmatrix}2\\0\end{bmatrix}.$$
But I am not able to see how the composition will be a rotation around the point $\big(1, -\tan \frac{\Theta}{2}\big).$ Kindly help me or tell me where I am wrong.
Look at this using affine transformations.
The rotation about the origin is
$$\small {\rm R}_{1}=\begin{bmatrix}\cos\Theta & \text{-}\sin\Theta\\ \sin\Theta & \cos\Theta\\ & & 1 \end{bmatrix}$$
and the rotation about $(2,0)$ is
$$\small {\rm R}_{2}=\begin{bmatrix}1 & & 2\\ & 1 & 0\\ & & 1 \end{bmatrix}\begin{bmatrix}\cos\Theta & \text{-}\sin\Theta\\ \sin\Theta & \cos\Theta\\ & & 1 \end{bmatrix}\begin{bmatrix}1 & & 2\\ & 1 & 0\\ & & 1 \end{bmatrix}^{-1}=\begin{bmatrix}\cos\Theta & \text{-}\sin\Theta & 2-2\cos\Theta\\ \sin\Theta & \cos\Theta & -2\sin\Theta\\ & & 1 \end{bmatrix}$$
You can confirm the above by multiplying the above with any coordinate point $\small \pmatrix{x \\ y \\ 1}$ and see if the results in the first two components match what you already have.
The combined transformation is
$$\begin{aligned}{\rm R}_{1}{\rm R}_{2} & =\begin{bmatrix}\cos\Theta & \text{-}\sin\Theta\\ \sin\Theta & \cos\Theta\\ & & 1 \end{bmatrix}\begin{bmatrix}1 & & 2\\ & 1 & 0\\ & & 1 \end{bmatrix}\begin{bmatrix}\cos\Theta & \text{-}\sin\Theta\\ \sin\Theta & \cos\Theta\\ & & 1 \end{bmatrix}\begin{bmatrix}1 & & -2\\ & 1 & 0\\ & & 1 \end{bmatrix}\\ & =\begin{bmatrix}\cos2\Theta & \text{-}\sin2\Theta & 2\cos\Theta-2\cos2\Theta\\ \sin2\Theta & \cos2\Theta & 2\sin\Theta-2\sin2\Theta\\ & & 1 \end{bmatrix} \end{aligned} $$
which is indeed a rotation of $2\Theta$. The center of rotation is checked against the point $\small \pmatrix{1 \\ \mbox{-}\tan \tfrac{\Theta}{2} }$ given by finding the transformation matrix about said point and checking with the above.
$$ \small \begin{bmatrix}1 & & 1\\ & 1 & \text{-}\tan\tfrac{\Theta}{2}\\ & & 1 \end{bmatrix}\begin{bmatrix}\cos2\Theta & \text{-}\sin2\Theta\\ \sin2\Theta & \cos2\Theta\\ & & 1 \end{bmatrix}\begin{bmatrix}1 & & 1\\ & 1 & \text{-}\tan\tfrac{\Theta}{2}\\ & & 1 \end{bmatrix}^{-1}\overset{\checkmark}{=}\begin{bmatrix}\cos2\Theta & \text{-}\sin2\Theta & 2\cos\Theta-2\cos2\Theta\\ \sin2\Theta & \cos2\Theta & 2\sin\Theta-2\sin2\Theta\\ & & 1 \end{bmatrix} $$
So the general form of rotation about a point $(t_x,t_y)$ is
$$\begin{aligned}{\rm R} & =\begin{bmatrix}1 & & t_{x}\\ & 1 & t_{y}\\ & & 1 \end{bmatrix}\begin{bmatrix}\cos\varphi & \text{-}\sin\varphi\\ \sin\varphi & \cos\varphi\\ & & 1 \end{bmatrix}\begin{bmatrix}1 & & \text{-}t_{x}\\ & 1 & \text{-}t_{y}\\ & & 1 \end{bmatrix}\\ & =\begin{bmatrix}\cos\varphi & \text{-}\sin\varphi & t_{x}-t_{x}\cos\varphi+t_{y}\sin\varphi\\ \sin\varphi & \cos\varphi & t_{y}-t_{y}\cos\varphi-t_{x}\sin\varphi\\ & & 1 \end{bmatrix}\\ & =\begin{bmatrix}\cos\varphi & \text{-}\sin\varphi & a\\ \sin\varphi & \cos\varphi & b\\ & & 1 \end{bmatrix} \end{aligned}$$
and given $a$, $b$ values from the 3rd column we extract the pivot with
$$\left.\begin{aligned}a & =t_{x}-t_{x}\cos\varphi+t_{y}\sin\varphi\\ b & =t_{y}-t_{y}\cos\varphi-t_{x}\sin\varphi \end{aligned} \right\} \begin{aligned}t_{x} & =\tfrac{1}{2}\,\frac{a-a\cos\varphi-b\sin\varphi}{1-\cos\varphi}\\ t_{y} & =\tfrac{1}{2}\,\frac{b-b\cos\varphi-a\sin\varphi}{1-\cos\varphi} \end{aligned}$$