Find the curve whose curvature is 2, passes through the point (1,0) and whose tangent vector at (1,0) is $\begin{bmatrix} 1/2\\ \sqrt{3}/2 \end{bmatrix}$. Hint: Use the proof of the fundamental theorem of plane curves.
So this is a differential geometry problem, but it feels like it can be done with tricks from trigonometry and calculus. Since the curvature is constant, and we are in $\mathbb{R}^2$, we know that the curve is a circle of radius 1/2. so say we have a parameterization like $\alpha(t) = \frac{1}{2}(-\cos(t),\sin(t)) + \vec{p} ~~~~~~ t \in [0,2\pi]$. Where $\vec{p}$ is our center point. We could get $T(t) = \alpha'(t) = \frac{1}{2}(\sin(t),\cos(t))$ but from here is where I am stuck. We could set this equal to our given tangent vector but what would this tell us about $t\in[0,2\pi]$? is this approach even any good? is $\vec{p}$ the only thing I need to define this curve?
EDIT: I have solved it using the hint, but I would still like to know if it is doable using elementary techniques.
To solve the problem elementarily, note that if the tangent vector is rotated 90° counterclockwise around $(1,0)$, so that it becomes $(-\frac{\sqrt3}2,\frac12)$, it points towards the circle's centre. This vector conveniently has length 1, so halving it produces a vector that points from $(1,0)$ to the circle centre. This centre is therefore $(1,0)+\frac12(-\frac{\sqrt3}2,\frac12)=(1-\frac{\sqrt3}4,\frac14)$.
The parametrisation of the required curve is thus $$\alpha(t)=\left(1-\frac{\sqrt3}4,\frac14\right)+\frac12(\cos t,\sin t),\ t\in[0,2\pi]$$