In one plane $Oxy$, given the circle $(\!{\rm C}\!): x^{2}+ y^{2}- 2x+ 4y- 4= 0$ around point $O$ for $60^{\circ}$ and it maps in the circle $(\!{\rm C}'\!)$. Find the loci of the circle $(\!{\rm C}'\!)$.
Firstly, we let ${\rm M}(x, y)\in (\!{\rm C}\!), {\rm M}'({x}',{y}')\in (\!{\rm C}'\!)$. We always have the following system of equations $$\left\{\begin{matrix} {x}'= x\cos\alpha- y\sin\alpha\\ {y}'= x\sin\alpha+ y\cos\alpha \end{matrix}\right.$$ or $$\left\{\begin{matrix} {x}'= x\cos 60^{\circ}- y\sin 60^{\circ}\\ {y}'= x\sin 60^{\circ}+ y\cos 60^{\circ} \end{matrix}\right.$$ so $$x= \frac{1}{2}{x}'+ \frac{\sqrt{3}}{2}{y}'$$ $$y= \frac{1}{2}{y}'- \frac{\sqrt{3}}{2}{x}'$$ Secondly https://www.wolframalpha.com/input/?i=(1%2F2a%2Bsqrt(3)%2F2b)%5E2%2B(-sqrt(3)%2F2a%2B1%2F2b)%5E2-2(1%2F2a%2Bsqrt(3)%2F2b)%2B4(-sqrt(3)%2F2a%2B1%2F2b)-4%3D0. So, how can I prove the given system of equations as follow ? $$\left\{\begin{matrix} x'= x\cos\alpha- y\sin\alpha\\ y'= x\sin\alpha+ y\cos\alpha \end{matrix}\right.$$ Finally, I need to the helps! Thanks for all the nice comments!
The question was:
Circle $\left ( C \right ): x^{2}+ y^{2}- 2\,x+ 4\,y- 4= 0$ $(1)$ in plane $Oxy$ around point $O$ for $60^\circ$ and it maps in circle $(C').$ Find $(C').$
Solution
$\left ( C \right ): x^{2}+ y^{2}- 2\,x+ 4\,y- 4= 0$ rewrites as $$(x-1)^2+(y+2)^2=9,$$ thus the center is $S(1,-2)$ and the radius $r=3.$ Through a rotation, the circle is transformed in a circle $(C')$ with equal radius $r=3$ and the center $$S'(1\cdot \cos 60^{\circ} -(-2)\sin 60^\circ,\,1\cdot \sin 60^\circ+(-2)\cos 60^\circ)=S'(1/2+\sqrt3,\sqrt3/2-1). $$
From this $\left ( C' \right ): \left(x-\frac{1+2\sqrt3}{2}\right)^2 + \left(y-\frac{\sqrt3-2}{2}\right)^2=9.$