[THOUGHT] Let me set up the coordinate system. I'll place the center of the circle at the origin, so the circle has radius $5\sqrt{2}$ (since the diameter is $10\sqrt{2}$).

Let me place $A$ and $B$ on the circle such that $AB$ is a diameter. I'll put:

• $A = (-5\sqrt{2}, 0)$
• $B = (5\sqrt{2}, 0)$

Since $P$ and $Q$ lie on the circle with diameter $AB$, they satisfy $x^2 + y^2 = 50$.

Given that $|PB| = 14$, I need to find the position of $P$ on the circle.

Let $P = (x_P, y_P)$ where $x_P^2 + y_P^2 = 50$.

$|PB|^2 = (x_P - 5\sqrt{2})^2 + y_P^2 = 196$

Expanding: $x_P^2 - 10\sqrt{2}x_P + 50 + y_P^2 = 196$

Since $x_P^2 + y_P^2 = 50$: $50 - 10\sqrt{2}x_P + 50 = 196$

$100 - 10\sqrt{2}x_P = 196$ $-10\sqrt{2}x_P = 96$ $x_P = -\frac{96}{10\sqrt{2}} = -\frac{48}{5\sqrt{2}} = -\frac{48\sqrt{2}}{10} = -\frac{24\sqrt{2}}{5}$

Then $y_P^2 = 50 - x_P^2 = 50 - \frac{576 \cdot 2}{25} = 50 - \frac{1152}{25} = \frac{1250 - 1152}{25} = \frac{98}{25}$

So $y_P = \pm\frac{\sqrt{98}}{5} = \pm\frac{7\sqrt{2}}{5}$

Let me take $P = (-\frac{24\sqrt{2}}{5}, \frac{7\sqrt{2}}{5})$ (we can check both cases).

Now I need to use the condition $(\overrightarrow{PA} + \overrightarrow{PB}) \cdot (\overrightarrow{PQ} + \overrightarrow{PB}) = 2|\overrightarrow{PQ}|^2$.

Let me compute this in Python. [/THOUGHT] [PYTHON] import sympy as sp from sympy import sqrt, symbols, solve, Rational, Abs

Set up coordinates

A = (-5sqrt(2), 0) B = (5sqrt(2), 0)

P on the circle with |PB| = 14

x_P = -24sqrt(2)/5 y_P = 7sqrt(2)/5 # Taking positive y first

P = (x_P, y_P)

Verify P is on the circle and |PB| = 14

print("P on circle:", sp.simplify(x_P2 + y_P2)) print("|PB|^2:", sp.simplify((x_P - 5*sqrt(2))2 + y_P2))

Q is on the circle: Q = (x_Q, y_Q) with x_Q^2 + y_Q^2 = 50

x_Q, y_Q = symbols('x_Q y_Q', real=True)

Vectors

PA = (A[0] - P[0], A[1] - P[1]) PB = (B[0] - P[0], B[1] - P[1]) PQ = (x_Q - P[0], y_Q - P[1])

PA + PB

PA_plus_PB = (PA[0] + PB[0], PA[1] + PB[1]) print("PA + PB:", sp.simplify(PA_plus_PB[0]), sp.simplify(PA_plus_PB[1]))

PQ + PB

PQ_plus_PB = (PQ[0] + PB[0], PQ[1] + PB[1])

Dot product (PA + PB) · (PQ + PB)

dot_product = PA_plus_PB[0] * PQ_plus_PB[0] + PA_plus_PB[1] * PQ_plus_PB[1] dot_product = sp.simplify(dot_product)

|PQ|^2

PQ_squared = PQ[0]**2 + PQ[1]**2 PQ_squared = sp.simplify(PQ_squared)

Condition: dot_product = 2 * PQ_squared

condition1 = sp.Eq(dot_product, 2 * PQ_squared) print("Condition 1:", condition1)

Circle condition for Q

condition2 = sp.Eq(x_Q2 + y_Q2, 50)

Solve for Q

solutions = solve([condition1, condition2], [x_Q, y_Q]) print("Solutions for Q:", solutions) [/PYTHON]

[THOUGHT] Now I have two solutions for Q:

1. Q = (527√2/125, -336√2/125)
2. Q = (5√2, 0) = B

Since |QB| > 0, Q ≠ B, so Q = (527√2/125, -336√2/125).

Now I need to compute |PA · QB|.

PA = A - P = (-5√2 - (-24√2/5), 0 - 7√2/5) = (-5√2 + 24√2/5, -7√2/5) = ((-25√2 + 24√2)/5, -7√2/5) = (-√2/5, -7√2/5)

QB = B - Q = (5√2 - 527√2/125, 0 - (-336√2/125)) = ((625√2 - 527√2)/125, 336√2/125) = (98√2/125, 336√2/125)

PA · QB = (-√2/5)(98√2/125) + (-7√2/5)(336√2/125) = (-√2 · 98√2)/(5·125) + (-7√2 · 336√2)/(5·125) = (-98·2)/625 + (-7·336·2)/625 = -196/625 - 4704/625 = -4900/625 = -196/25

So |PA · QB| = 196/25.

Since 196 = 4·49 = 4·49 and 25 = 25, gcd(196, 25) = 1.

So q = 196, p = 25, and p + q = 221.

Wait, let me verify: the problem says |PA · QB| = q/p where p and q are coprime natural numbers. So 196/25 means q = 196, p = 25.

p + q = 25 + 196 = 221. [/THOUGHT]

\boxed{221}