How to translate to a specific point with rotational transformation.

Question

Basically I have two rectangles. ABCD and EFGH

EFGH is rotated around it's centre point (X)

ABCD has centre point (W)

I also know for the sake of this example that EFGH is rotated counter clockwise at 45°

I am trying to do this for all corners but basically

What I am trying to figure out is how much do I need to translate ABCD by such that when rotated at 45° around its centre point (W), the corner B will be equal to point F. (See blue square as a rough example)

Basically Im trying to solve for what coordinates does the centre W have to be such that after rotation around it B is equal to F

I've spent ages trying to figure this out and I can't come up with anything. I'm not too familiar with matrices.

I tried calculating the distance and angle of WB, and since I know what F is i could say

x = WB * cos (WB°+45°)
y = WB * sin (WB°+45°)

But Fx = x + x2(origin)
And Fy = y + y2(origin)

x2 = Fx - x
y2 = Fy-y

Trying to solve for new 'origin' of ABCD, but to no avail..

Answer 1

Let $X=(0,0)$ and let $W=(a,b)$. Call $S_X, S_W$ the respective squares, $S_X', S_W'$ after rotation.

$$S_X : (a\pm EF/2, b\pm EF/2)$$

$$S_W : (\pm AB/2, \pm AB/2)$$

After rotation:

$$S_X' : (a, b\pm \sqrt{2}/2EF),(a\pm \sqrt{2}/2EF, b)$$

$$S_W' : (0, \pm \sqrt{2}/2AB), (\pm \sqrt{2}/2AB,0)$$

(you can get this by multiplying the coordinates by the rotation matrix.)

If you want to line up $B$ and $F$ then you want real numbers $s,t$ such that

$$(s, \sqrt{2}/2AB+t)=(a, b+ \sqrt{2}/2EF)$$

So $$(s,t)= (a,b+\sqrt{2}/2(EF-AB))$$

Edit

Here is an interactive plot I made for you. First both squares of specified length are rotated by 45 degrees. Then the square centered at the origin is translated by the amount specified above.

https://www.desmos.com/calculator/psp0oquhfs

For rectangles, rather than go through a whole 'nother derivation, the value of $s,t$ you seek is:

$$\left[\begin{matrix}s\\ t\end{matrix}\right]=\left[\begin{matrix}a\\ b\end{matrix}\right]+\left[\begin{matrix}\cos\theta & -\sin\theta\\ \sin\theta & \cos\theta\end{matrix}\right]\left[\begin{matrix}F_x-B_x-a\\ F_y-B_y-b\end{matrix}\right]$$

Another example for rectangles, rotating $270^o$

https://www.desmos.com/calculator/qccjkp32ha