[THOUGHT] We need to find the time (t_0) in ([0,2\pi]) where the distance between the two moving points is maximized.
The points are: [ P_1(t) = (1+\cos t,;\sin t),\qquad P_2(t) = (2\cos\theta(t),;2\sin\theta(t)). ] Given the speed ratio (|P_1'(t)|:|P_2'(t)| = 1:4), we have [ |P_1'(t)| = 1,\qquad |P_2'(t)| = 4. ] Since (|P_2'(t)| = 2|\theta'(t)|), we get (|\theta'(t)| = 2).
Both start at ((2,0)) at (t=0), so (\theta(0)=0) and (\theta(t)=2t).

The squared distance is [ D^2(t) = (1+\cos t-2\cos 2t)^2 + (\sin t-2\sin 2t)^2. ] Simplifying with (\cos 2t = 2\cos^2 t-1) and (\sin 2t = 2\sin t\cos t) yields [ D^2(t) = -8\cos^2 t - 2\cos t + 10. ] Differentiating, [ \frac{dD^2}{dt}=2\sin t,(8\cos t+1)=0, ] so critical points are (\sin t=0) (giving (D^2=0) or (4)) and (\cos t=-\frac18).
The latter gives the maximum distance: [ t_0 = \arccos!\left(-\frac18\right). ]

The required expression is [ E(t)=\sin t\cdot2\sin\theta(t)-(1+\cos t)\cdot2\cos\theta(t) =2\bigl[\sin t\sin 2t-(1+\cos t)\cos 2t\bigr]. ] Using (\sin 2t = 2\sin t\cos t) and (\cos 2t = 2\cos^2 t-1) we simplify to [ E(t) = -4\cos^2 t - 8\cos^3 t + 6\cos t + 2. ] Substituting (\cos t_0 = -\frac18): [ E(t_0)= -\frac{4}{64} -\frac{8}{512} -\frac{3}{4} + 2 = \frac{77}{64}. ] Thus (p=77,;q=64) and (p+q=141).

We will verify the calculations with Python. [/THOUGHT]

[PYTHON] import sympy as sp

Define symbol

t = sp.symbols('t', real=True)

Define cos and sin

c = sp.cos(t) s = sp.sin(t)

Distance squared

D2 = (1 + c - 2sp.cos(2t))**2 + (s - 2sp.sin(2t))**2 D2_simplified = sp.simplify(D2) print("D2 simplified:", D2_simplified)

Derivative for critical points

dD2 = sp.diff(D2_simplified, t) critical_eq = sp.Eq(dD2, 0) critical_points = sp.solveset(dD2, t, domain=sp.Interval(0, 2*sp.pi)) print("Critical points:", critical_points)

Choose the non-trivial maximum (cos t = -1/8)

t0 = sp.acos(-sp.Rational(1,8)) print("t0:", t0)

Verify D2 at t0

D2_at_t0 = D2_simplified.subs(t, t0).simplify() print("D2 at t0:", D2_at_t0)

Expression E

E = s2sp.sin(2t) - (1 + c)2sp.cos(2t) E_simplified = sp.simplify(E) print("E simplified:", E_simplified)

Evaluate E at t0

E_at_t0 = E_simplified.subs(t, t0).simplify() print("E at t0:", E_at_t0) [/PYTHON]

\boxed{141}

[THOUGHT] The Python output confirms the expression simplifies to ( E(t_0) = \frac{31}{16} - 2\cos(3t_0) ) where ( t_0 = \arccos(-1/8) ). Using the triple angle formula ( \cos(3t) = 4\cos^3 t - 3\cos t ) with ( \cos t_0 = -1/8 ), we compute: [ \cos(3t_0) = 4\left(-\frac{1}{8}\right)^3 - 3\left(-\frac{1}{8}\right) = -\frac{1}{128} + \frac{3}{8} = \frac{47}{128}. ] Substituting: [ E(t_0) = \frac{31}{16} - 2 \cdot \frac{47}{128} = \frac{248}{128} - \frac{94}{128} = \frac{154}{128} = \frac{77}{64}. ] Thus, ( p = 77 ), ( q = 64 ), and ( p + q = 141 ). [/THOUGHT]

\boxed{141}