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 $S(t)$:

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 $t = t_0$:

Substitute $\sin\theta = \sin2t = 2\sin t\cos t$ and $\cos\theta = \cos2t = 2\cos^2 t-1$:

Using $\sin^2\!t = 1 - \cos^2\!t$ and simplifying yields an expression solely in $c = \cos t$:

Now set $c = -\frac{1}{8}$:

Adding with denominator $64$:

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 $p+q = 77+64 = 141$.

$\boxed{141}$