Let $C = (a, b)$ be any given point in the plane, and let $r$ be any given positive real number. Then how to find the coordinates of the vertices of an equilateral triangle inscribed in the circle $$ \left\{ (x, y) \in \mathbb{R}^2 \colon (x-a)^2 + (y-b)^2 = r^2 \right\}? $$
My Attempt:
Let $P$, $Q$, $R$ be the vertices one such equilateral triangle. Then we have $$ \triangle PCQ \cong \triangle QCR \cong \triangle RCP. $$ So we have $$ \angle PCQ \cong \angle QCR \cong \angle RCP, $$ which implies that $$ m \angle PCQ = m \angle QCR = m \angle RCP = \frac{\pi}{3}. $$ Moreover, we also have $$ \vec{OP} = \vec{OC} + \vec{CP}. $$ If $\alpha$ is the angle that $\vec{CP}$ makes with the positive $x$-axis, where $0 \leq \alpha < 2 \pi$, then we have $$ \vec{OP} = (a + r \cos \alpha ) \hat{i} + ( b + r \sin \alpha ) \hat{j}. $$ Thus we have $$ P = \big( a + r \cos \alpha, b + r \sin \alpha \big). $$ Since the angle between the vectors $\vec{CP}$ and $\vec{CQ}$ and the angle between $\vec{CQ}$ and $\vec{CR}$ is $\frac{\pi}{3}$, we have $$ \vec{OQ} = \vec{OC} + \vec{CQ} = \left( a + r \cos \left( \alpha + \frac{\pi}{3} \right) \right) \hat{i} + \left( b + r \sin \left( \alpha + \frac{\pi}{3} \right) \right) \hat{j}, $$ and therefore we have $$ Q = \left( a + r \cos \left( \alpha + \frac{\pi}{3} \right) \, , \, b + r \sin \left( \alpha + \frac{\pi}{3} \right) \right). $$ Similarly, we have $$ R = \left( a + r \cos \left( \alpha + \frac{2\pi}{3} \right) \, , \, b + r \sin \left( \alpha + \frac{2 \pi }{3} \right) \right). $$
Is my solution correct and clear enough? Or, are there any problems?