How to construct an equilateral triangle on 2 concentric circles 
Construct an equilateral triangle with the given vertex so that the other vertices lie on the concentric circles respectively. 

I constructed the triangle, but I don't  know how it works. How does this construction work?  Is there any proof?  
My construction. 
Let the smaller circle be $a$, the larger circle $b$, and  the point $c$.     


*

*Step 1:  Construct a circle with radius of $b$ at the point $c$.  

*Step 2: The circle will intersect circle $a$ at  $2$ points.  Let the two points be $x$ and $y$.   Construct a perpendicular bisector of the line connecting $x$ and the common centre of circle $a$ and $b$.  

*Step 3:  The bisector intersect the circle $a$ at a point which is another vertex of the equilateral triangle.

For more context, this is from the game "Euclidea" level 13.3.  Video solutions can be found here.
 A: 
Task.  Given a point $P$ on the plane and two (not necessarily distinct and not necessarily concentric) circles $c$ and $k$, construct an equilateral $PAB$ such that $A$ is a point of $c$ and $B$ is a point of $k$.


Construction.  Denote by $c'$ and $k'$ the images of $c$ and $k$, respectively, under the counterclockwise rotation about $P$ by $\dfrac{\pi}{3}$.  Suppose that $c$ meets $k'$ at $A$ and $A'$, and that $c'$ meets $k$ at $B''$ and $B'''$.  Let $B$, $B'$, $A''$, and $A'''$ be the images of $A$, $A'$, $B''$, and $B'''$ under the clockwise rotation about $P$ by $\dfrac{\pi}{3}$.  Then, $PAB$, $PA'B'$, $PA''B''$, and $PA'''B'''$ are equilateral triangles.  The number of such triangles can be $0$, $1$, $2$, $3$, and $4$, depending how $c$ and $k$ intersect $c'$ and $k'$.

Explaination.  If $PAB$ is a desired triangle, then $A$ is the image of counterclockwise rotation about $P$ by $\theta\in\left\{-\dfrac{\pi}{3},+\dfrac{\pi}{3}\right\}$.  If $\theta=+\dfrac{\pi}{3}$, then clearly, $A$ is a point of intersection between $c$ and $k'$.  If $\theta=-\dfrac{\pi}{3}$, then $B$ is the point of intersection between $c'$ and $k$.

Addendum.
The OP's construction works when the two circles are concentric.  I have not yet found out why.  If I know the answer, I will come back to give a proof.  For now, I attach a figure showing that the OP's steps do lead to a correct construction.

A: As shown in one of the other answers, to find another vertex of the equilateral triangle one can rotate the outer circle $b$ by $60°$ about the given vertex $P$: each intersection between the rotated circle and the inner circle $a$ is then a possible second vertex of the equilateral triangle to be constructed.
The construction you found works because, instead of rotating $b$ about $P$ by $60°$ counterclockwise, we can obtain the same result by rotating a circle equal to $b$ but centred at $P$ by $60°$ clockwise about the common center $O$ of $a$ and $b$.
In figure below, the red circle is obtained by rotating circle $b$ about $P$ by $60°$ counterclockwise; its intersection $A'$ with circle $a$ is the second vertex of the equilateral triangle to be constructed.
But we can also find the red circle by constructing first a circle equal to $b$ centred at $P$ (blue circle in the figure) and then rotating it about $O$ by $60°$ clockwise. Point $A'$ can then be quickly obtained by rotating point $A$ (the intersection between $a$ and blue circle) by $60°$ clockwise about $O$: as triangle $AOA'$ is equilateral, $A'$ is thus the intersection between $b$ and the perpendicular bisector of $OA$.

A: COMMENT.-An easily verifiable fact is that given two points $Q, R$  one in each circle, there is always a point $P$ such that the triangle $\triangle PQR$ is equilateral, but another problem is the one in which the point $P$ occurs and the points to be determined are $Q$ and $R$. Look just at first glance an analytical solution.
Being $R$ and $r$ the radius and the point $P=(a,0)$ we have for the two circles and the three equal distances the four equations determining the points $Q=(x,y)$ and $R=(w,z)$ in both circles
$$x^2+y^2=r^2\\z^2+w^2=R^2\\(x-a)^2+y^2=(z-a)^2+w^2=(x-z)^2+(y-w)^2$$
We have $z=B+x$ where $B=\dfrac{R^2-r^2}{2a}$ and $w=\dfrac{C+2Dx-2x^2}{2y}$ where $C=R^2-a^2$ and $D=a-B$. Then $$x^2+y^2=r^2\\(x+B)^2+\left(\dfrac{C+2Dx-2x^2}{2y}\right)^2=R^2$$ so the resultant, where the coefficients $c_i$ are constant $$c_1x^3+c_2x^2+c_3x+c_4=0$$.
This equation has always a real root but this does not guarantee that there is always a solution to the problem posed. In fact, the smallest and greatest possible distance between two points, one in each circle, is $R-r$ and $R + r$ respectively. Consequently, for every point outside the interior of the circle of radius $2R+r$ we can assure that there is no solution.
