= Solution
{c}
A possible center $\mathbf c$ must be a <fixed point>, so $(I-R_\theta)\mathbf c=\mathbf b$. Let $c=\cos\theta$ and $s=\sin\theta$. The <determinant> is
$$
\det(I-R_\theta)=(1-c)^2+s^2=2(1-c)=4\sin^2(\theta/2)>0
$$
when $\theta$ is not an <integer multiple> of $2\pi$. The unique center is therefore
$$
\boxed{\mathbf c=(I-R_\theta)^{-1}\mathbf b=\frac1{2(1-\cos\theta)}\begin{pmatrix}1-\cos\theta&-\sin\theta\\\sin\theta&1-\cos\theta\end{pmatrix}\mathbf b.}
$$
The <fixed point> equation implies $\mathbf b=\mathbf c-R_\theta\mathbf c$, whence
$$
T(\mathbf v)=\mathbf c+R_\theta(\mathbf v-\mathbf c).
$$
Relative to $\mathbf c$, this is precisely a <planar rotation> through $\theta$. This also explains the excluded case: when $\theta\in2\pi\mathbb Z$, $T$ is a <translation>, with no <fixed point> unless $\mathbf b=0$.
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