We have the two moving points:

• (P_1(t) = (1 + \cos t,; \sin t)) on the circle centered at ((1,0)) radius (1).
• (P_2(t) = (2\cos\theta(t),; 2\sin\theta(t))) on the circle centered at the origin radius (2).

Both start at ((2,0)) and move counter‑clockwise.
The speed condition is (|P_1'(t)| : |P_2'(t)| = 1 : 4) for (0 \le t \le 2\pi).
Since (|P_1'(t)| = 1) (constant), we have (|P_2'(t)| = 4).
Now (P_2'(t) = 2\theta'(t)(-\sin\theta,\cos\theta)), so (|P_2'(t)| = 2|\theta'(t)|).
Hence (2|\theta'(t)| = 4 \Rightarrow |\theta'(t)| = 2). With counter‑clockwise motion, (\theta'(t) = 2), and (\theta(0)=0) gives (\theta(t) = 2t).
Thus (P_2(t) = (2\cos 2t,; 2\sin 2t)).

We need the time (t_0) when the distance (D(t) = |P_1(t)-P_2(t)|) is maximised.
Because the square root is increasing, we maximise (S(t) = |P_1-P_2|^2).

Compute (S(t)): [ \begin{aligned} S(t) &= (1+\cos t - 2\cos 2t)^2 + (\sin t - 2\sin 2t)^2 \ &= 6 + 2\cos t - 4\bigl((1+\cos t)\cos 2t + \sin t\sin 2t\bigr) \ &= 6 + 2\cos t - 4(\cos 2t + \cos t) \qquad\text{(using the product‑to‑sum identity)} \ &= 6 + 2c - 4\bigl((2c^2-1) + c\bigr) \qquad (c = \cos t,; \cos 2t = 2c^2-1) \ &= 10 - 2c - 8c^2. \end{aligned} ]

Thus (S(t) = -8c^2 - 2c + 10) with (c = \cos t). This is a downward‑opening quadratic in (c) with vertex at (c = -\frac{1}{8}).
Since (-\frac{1}{8}\in[-1,1]), the maximum of (S(t)) occurs when (\cos t = -\frac{1}{8}).
Hence (t_0) satisfies (\cos t_0 = -\frac{1}{8}). (There are two such times in ([0,2\pi]), but the expression to evaluate depends only on (\cos t_0).)

Now evaluate the given expression at (t = t_0): [ E = 2\sin t \cdot \sin\theta(t) - (1+\cos t)\cdot 2\cos\theta(t), \qquad \theta(t)=2t. ] Substitute (\sin\theta = \sin2t = 2\sin t\cos t) and (\cos\theta = \cos2t = 2\cos^2 t-1): [ \begin{aligned} E &= 2\sin t\sin2t - 2(1+\cos t)\cos2t \ &= 4\sin^2!\cos t - 2(1+\cos t)(2\cos^2!t-1). \end{aligned} ] Using (\sin^2!t = 1 - \cos^2!t) and simplifying yields an expression solely in (c = \cos t): [ E = -8c^3 - 4c^2 + 6c + 2. ]

Now set (c = -\frac{1}{8}): [ \begin{aligned} c^2 = \frac{1}{64},\qquad c^3 = -\frac{1}{512}, \[2mm] -8c^3 &= -8!\left(-\frac{1}{512}\right) = \frac{1}{64},\[2mm] -4c^2 &= -4!\left(\frac{1}{64}\right) = -\frac{4}{64} = -\frac{1}{16},\[2mm] 6c &= 6!\left(-\frac{1}{8}\right) = -\frac{6}{8} = -\frac{48}{64},\[2mm] +2 &= \frac{128}{64}. \end{aligned} ] Adding with denominator (64): [ \frac{1}{64} - \frac{4}{64} - \frac{48}{64} + \frac{128}{64} = \frac{77}{64}. ] Thus (E = \frac{77}{64}), so (p = 77), (q = 64) (coprime).
Therefore (p+q = 77+64 = \boxed{141}).

[THOUGHT] We need to compute the maximum of distance squared between P1(t) and P2(t), find t0, and evaluate the given expression. Let's do it step by step with Python.

We know: P1(t) = (1 + cos t, sin t) P2(t) = (2 cos 2t, 2 sin 2t) [since θ = 2t]

Distance squared: S(t) = (1 + cos t - 2 cos 2t)^2 + (sin t - 2 sin 2t)^2.

We can simplify symbolically or evaluate numerically. We expect maximum at cos t = -1/8.

We'll compute S(t) for t in [0, 2π], find t0 where S is max, then compute E = 2 sin t sin(2t) - 2(1+cos t) cos(2t) at t0.

Let's code. [/THOUGHT]

[PYTHON] import numpy as np

Define the distance squared function

def S(t): return (1 + np.cos(t) - 2np.cos(2t))**2 + (np.sin(t) - 2np.sin(2t))**2

Sample many points to find approximate t0

t_vals = np.linspace(0, 2*np.pi, 1000001) S_vals = S(t_vals) max_idx = np.argmax(S_vals) t0 = t_vals[max_idx] print("t0 (approx):", t0) print("cos(t0):", np.cos(t0))

Now compute E at t0

c = np.cos(t0)

Using exact expression: E = -8c^3 - 4c^2 + 6c + 2

E = -8c**3 - 4c**2 + 6*c + 2 print("E (approx):", E)

Alternatively, compute directly from trig:

theta = 2t0 E_direct = 2np.sin(t0)np.sin(theta) - 2(1+np.cos(t0))*np.cos(theta) print("E_direct (approx):", E_direct) </PYTHON>

[THOUGHT] Let's redo the code correctly.

[PYTHON] import numpy as np

Define the distance squared function

def S(t): return (1 + np.cos(t) - 2np.cos(2t))**2 + (np.sin(t) - 2np.sin(2t))**2

Sample many points to find approximate t0

t_vals = np.linspace( univers0.52, 1)

We'll sample t from 0 to 2π.

Let's compute: </think>

[THOUGHT] Let's redo the code carefully.

[PYTHON] import numpy as np

Define the distance squared function

def S(t): return (1 + np.cos(t) - 2np.cos(2t))**2 + (np.sin(t) - 2np.sin(2t))**2

Sample many points to find approximate t0

t_vals = np.linspace(0, 2*np.pi, 1000001) S_vals = S(t_vals) max_idx = np.argmax(S_vals) t0 = t_vals[max_idx] print("t0 (approx):", t0) print("cos(t0):", np.cos(t0))

Now compute E at t0

c = np.cos(t0)

Using exact expression: E = -8c^3 - 4c^2 + 6c + 2

E = -8c**3 - 4c**2 + 6*c + 2 print("E (approx):", E)

Alternatively, compute directly from trig:

theta = 2t0 E_direct = 2np.sin(t0)np.sin(theta) - 2(1+np.cos(t0))*np.cos(theta) print("E_direct (approx):", E_direct) [/PYTHON]

[TH]

[THOUGHT] We compute E exactly using fractions to verify p/q = 77/64.

[PYTHON] from fractions import Fraction c = Fraction(-1, 8) E = -8c**3 - 4c**2 + 6*c + 2 print(E) print(E.numerator, E.denominator) print("p+q =", E.numerator + E.denominator) [/PYTHON] </think>

[THOUGHT] We compute E exactly using fractions to verify p/q = 77/64.

[PYTHON] from fractions import Fraction c = Fraction(-1, 8) E = -8c**3 - 4c**2 + 6*c + 2 print(E) print("p+q =", E.numerator + E.denominator) [/PYTHON]

[THOUGHT]
The expression evaluates to ( \frac{77}{64} ), with (p=77) and (q=64) relatively prime. Therefore (p+q = 77+64 = 141).

\boxed{141}