First of all, there may be multiple rectangles that satisfy your conditions, e.g.
if you want specific angle of rectangle, or
if you want to specify a point on the boundary.
However, there is a special case where there is at most only once rectangle, i.e. if you assume that it has to touch both of your boundaries. In such case it is easy to compute it. Place the origin (point $(0,0)$) in the center of the boundary (the center of the circle), and let $(C_x,C_y)$ be the top right boundary corner (the boundary rectangle has size $2C_x \times 2C_y$ ). Let $(x,y)$ be the top right vertex of small rectangle, then it satisfies conditions ($\alpha > 0$ means counterclockwise rotation):
\begin{align*}
\left[\begin{matrix}x'\\\ C_y\end{matrix}\right] &= \left[\begin{matrix}\cos\alpha&-\sin\alpha\\\sin\alpha&\cos\alpha\end{matrix}\right]\left[\begin{matrix}x\\\y\end{matrix}\right] \\\
\left[\begin{matrix}C_x\\\ y'\end{matrix}\right] &= \left[\begin{matrix}\cos\alpha&-\sin\alpha\\\sin\alpha&\cos\alpha\end{matrix}\right]\left[\begin{matrix}x\\\ -y\end{matrix}\right]
\end{align*}
where $x'$ and $y'$ are just placeholders. Extracting appropriate rows from those formulae, we can transform that into one equation (notice the lack of minus sign in the matrix):
\begin{align*}
\left[\begin{matrix}C_x\\\ C_y\end{matrix}\right] &= \left[\begin{matrix}\cos\alpha&\sin\alpha\\\sin\alpha&\cos\alpha\end{matrix}\right]\left[\begin{matrix}x\\\ y\end{matrix}\right]
\end{align*}
with solution being:
\begin{align*}
\left[\begin{matrix}\cos\alpha&\sin\alpha\\\sin\alpha&\cos\alpha\end{matrix}\right]^{-1}
\left[\begin{matrix}C_x\\\ C_y\end{matrix}\right] &= \left[\begin{matrix}x\\\ y\end{matrix}\right] \\\
\sec{2\alpha}\left[\begin{matrix}\cos\alpha&-\sin\alpha\\\ -\sin\alpha&\cos\alpha\end{matrix}\right]
\left[\begin{matrix}C_x\\\ C_y\end{matrix}\right] &= \left[\begin{matrix}x\\\ y\end{matrix}\right]
\end{align*}
Please note, that this may not have a proper solution if $\alpha$ is to big!
Edit: Ok, I missed the comment about maximizing the area. Then again, consider this example:
The gray figure is a rhombus (it was created by rotating the black rectangle by $2\alpha$ ). To get the inscribed rectangle with the greatest area, consider the case when the rhombus would be a square--the greatest area would be when each rectangle vertex splits the rhombus edge in half (because then it is also a square and that is the rectangle with greatest area and given perimeter).
But we can scale our rhombus (that may not be a square) so that is a square, obtain the solution there, and then scale back (the area will scale accordingly)! In conclusion the rectangle of greatest area will split the rhombus edges in half.
How to compute it? You could do it using the same approach:
\begin{align*}
\left[\begin{matrix}x'\\\ C_y\end{matrix}\right] &= \left[\begin{matrix}\cos\alpha&-\sin\alpha\\\sin\alpha&\cos\alpha\end{matrix}\right]\left[\begin{matrix}x\\\y\end{matrix}\right] \\\
\left[\begin{matrix}x''\\\ -C_y\end{matrix}\right] &=
\left[\begin{matrix}\cos(-\alpha)&-\sin(-\alpha)\\\sin(-\alpha)&\cos(-\alpha)\end{matrix}\right]
\left[\begin{matrix}x\\\ -y\end{matrix}\right]
\end{align*}
However, one can do it simpler: the vertex of the inscribed rectangle splits the
gray edge in half, so $$2x\sin\alpha = C_y = 2y\cos\alpha\,.$$ This works if $C_x > C_y$, otherwise you need to do the same for $C_x$ instead. Moreover, even if $C_x > C_y$, you still need to check if the rotated rectangle fits into the boundary (because it may be that $C_x = C_y + \varepsilon $ ), if not, the solution from previous part will do.
Hope that helps ;-)
To solve this problem, you need to first calculate the coordinates of the corners of your rectangle. Let's name the corners of this rectangle as $MNPQ$. When the angle $\alpha=0$, the coordinate of point $M$ with respect to the center of the rectangle $(x_c,y_c)$ is $(-w/2,h/2)$. The angle that this point is seen from the center of the rectangle is $\gamma_M=\arctan2(-w/2,h/2)=\arctan2(-w,h)$, and the distance to the center is $\sqrt{w^2+h^2}/2$. In terms of polar coordinates the rotation by $\alpha$ means that we just need to add $\alpha$ to this $\gamma_M$. If you keep the trigonometric sense for the angles then the $\alpha$ in your figure is negative. We also need to add the coordinates of the center of the rectangle to this value, in order to get the coordinates in the original reference frame. What you get in this case is:
$$x_M=x_c+\cos(\gamma_M+\alpha)\sqrt{w^2+h^2}/2\\y_M=y_c+\sin(\gamma_M+\alpha)\sqrt{w^2+h^2}/2$$
Similarly you can get the coordinates of all corners using
$$\gamma_M=\arctan2(-w,+h)\\
\gamma_N=\arctan2(+w,+h)\\
\gamma_P=\arctan2(+w,-h)\\
\gamma_Q=\arctan2(-w,-h)$$
The equation of the line $MN$ can be written as$$\frac{x-x_M}{x_N-x_M}=\frac{y-y_M}{y_N-y_M}$$
while the equation of $OA$ line is $$\beta=\arctan2(y,-x)$$ I assumed here that the $\beta$ is measured clockwise from the $-y$ axis. You should modify this equation if it's not the case.
From the last two equations we can find the intersection of $OA$ with $MN$, meaning the coordinates of point $A$ if the intersection is between $M$ and $N$. That means that it's a solution if $(x_A-x_M)(x_A-x_N)\le0$. Similarly, you should find intersection points with $NP, PQ, QM$. You might find 0, 1, or 2 intersections
Best Answer
The following figure depicts the "original" arrangement when both rectangles are placed in the same coordinate system (red) with their centers in the origin:
Turn, about it center, the blue triangle counterclockwise until its upper-right corner meets the bigger rectangle. The angle at which this takes place is the same angle by which we would have turned the white rectangle clockwise to hit the blue rectangle.
Take the red circle centered at the origin and calculate its radius: $r=\frac12\sqrt{a_x^2+a_y^2}.$ So, the equation of the red circle is $$x^2+y^2=\frac14\left(a_x^2+a_y^2 \right).$$
Where does this circle hits the straight $y=y_b$? Now, you have two points: the upper right corner of the blue rectangle and the crossing point mentioned above. Can you calculate the angle between the two red lines?