Richard sent us his work on this problem.

We can say that $A P S$ and $B Q P$ are similar triangles (ratio of sides same), so in terms of angles:
$∠ A P S = ∠ B Q P$
$= 180^{\circ} - 90^{\circ} - ∠ B P Q$ (angles in triangle add to $180^{\circ}$ )
$= 90^{\circ} - ∠ B P Q$
$∠ S P Q + ∠ A P S + ∠ B P Q = 180^{\circ}$ (angles from a point on a straight line add to $180^{\circ}$ )
$\Rightarrow ∠ S P Q + (90^{\circ} - ∠ B P Q) + ∠ B P Q = 180^{\circ}$
$\Rightarrow ∠ S P Q + 90^{\circ} = 180^{\circ}$
$\Rightarrow ∠ S P Q = 90^{\circ}$ .
Same applies again here, triangle $Q C R$ is similar to $R D S$ , and so in terms of angles:
$∠ C R Q = ∠ D S R$
$= 90^{\circ} - ∠ S R D$
$∠ S R Q + ∠ C R Q + ∠ S R D = 180^{\circ}$
$∠ S R Q + (90^{\circ} - ∠ S R D) + ∠ S R D = 180^{\circ}$
$∠ S R Q = 90^{\circ}$ .
As two opposite angles add up to $180^{\circ}$ , the other two must as well (angles in quadrilateral add up to $360^{\circ}$ ). Two pairs of opposite angles each adding to $180^{\circ}$ implies a cyclic quadrilateral (one of the Circle Theorems).
Firstly, as $∠ S P Q$ and $∠ S R Q$ are both $90^{\circ}$ , line $S Q$ must be the diameter of the circle with midpoint being centre of circle, $M$ .
As $A P R D$ is a rectangle, $∠ A P R$ and $∠ D R P$ are both right angles. Also lines $P M = R M$ , as $P$ and $R$ are on the circle.
Midpoint $M$ is also the centre of the rectangle. Let $V$ be the length of the perpendicular from $M$ to side $A B$ , and let $U$ be the length from the foot of this perpendicular to $P$ , as shown in the diagram.

$U = [y (x - y) + x (x + y)] / 2 - A P$
$= [x y - y^{2} + x^{2} + x y] / 2 - y (x - y)$
$= [x^{2} + y^{2}] / 2$
$V = [y (x + y) + x (x - y)] / 2$
$= [x y + y^{2} + x^{2} - x y] / 2$
$= [x^{2} + y^{2}] / 2$
As $U = V$ , $∠ B P M = 45^{\circ}$
$\Rightarrow ∠ R P M = 45^{\circ}$
$\Rightarrow ∠ P R M = 45^{\circ}$ (isosceles triangle)
$\Rightarrow ∠ P M R = 180^{\circ} - 45^{\circ} - 45^{\circ} = 90^{\circ}$
$\Rightarrow ∠ P Q R = 45^{\circ}$ (angle at centre is double that at edge - Circle Theorem)
As $∠ P Q R = 45^{\circ}$
$\Rightarrow ∠ P S R = 180^{\circ} - 45^{\circ}$ (Cyclic Quadrilateral)
$= 135^{\circ}$ .