[THOUGHT] We need to understand the problem step by step.

Given: Point A(-2,0). Points P and Q satisfy: (i) Vectors AP and AQ are parallel. That means they are collinear with A, so P and Q lie on the same line through A. (ii) Dot product AP · AQ = 9.

Let AP = vector from A to P, AQ = vector from A to Q. Since they are parallel, we can write AQ = t * AP for some scalar t (real). Then AP · AQ = AP · (t AP) = t |AP|^2 = 9.

Thus t = 9 / |AP|^2.

Given P, Q = A + t * (P - A) = A + (9 / |AP|^2) * (P - A).

So Q is determined by P via a mapping: Q = f(P) = A + (9 / |AP|^2) * (P - A).

Now P moves along region R. Region R is described: circles C1: x^2 + y^2 = 1, C2: (x+1)^2 + y^2 = 1 intersect at two distinct points R1 and R2. The region R consists of arc R1 O' R2 of C1 containing O'(-1,0) together with arc R1 O R2 of C2 containing O(0,0). So R is a closed curve formed by two arcs: one from C1 (the arc that contains O'(-1,0)) and one from C2 (the arc that contains O(0,0)). The arcs share endpoints R1 and R2.

We need to find the total length of the locus traced by Q as P moves along R. That is, the length of the image curve f(R). So we need to compute the length of the curve Q(t) = f(P(t)) as P traverses R.

We can parametrize each arc separately and compute the length of the image under f.

First, find intersection points R1, R2 of circles C1 and C2.

C1: x^2 + y^2 = 1. C2: (x+1)^2 + y^2 = 1 => x^2 + 2x + 1 + y^2 = 1 => x^2 + y^2 + 2x = 0.

Substitute y^2 = 1 - x^2 from C1 into C2: (1 - x^2) + x^2 + 2x = 0 => 1 + 2x = 0 => x = -1/2.

Then y^2 = 1 - (1/4) = 3/4 => y = ±√3/2.

Thus R1 = (-1/2, √3/2), R2 = (-1/2, -√3/2).

Now region R: arcs:

Arc on C1: from R1 to R2 passing through O'(-1,0). Since O' is at (-1,0) on C1? Check: (-1)^2 + 0^2 = 1, yes O' is on C1. So the arc on C1 containing O' between R1 and R2 is the longer arc? Let's see: On circle C1, points R1 and R2 have x = -1/2, y = ±√3/2. The center of C1 is (0,0). The angle of R1: cosθ = x = -1/2, sinθ = √3/2 => θ = 2π/3 (120°). R2: θ = -2π/3 (or 4π/3). O'(-1,0) corresponds to angle π (180°). So the arc from R1 (120°) to R2 (240°) going through 180° is the arc from 120° to 240° (counterclockwise). That's length of arc: central angle 120° = 2π/3, radius 1, so arc length = 2π/3.

Arc on C2: from R1 to R2 passing through O(0,0). C2 center is (-1,0). O(0,0) is on C2? (0+1)^2 + 0^2 = 1, yes. Coordinates of R1 relative to center (-1,0): R1 - (-1,0) = (-1/2 +1, √3/2) = (1/2, √3/2). That's angle 60° (π/3). R2: (1/2, -√3/2) angle -60° (or 300°). O(0,0) relative to center (-1,0) is (1,0) angle 0°. So the arc from R1 (60°) to R2 (300°) passing through 0° is the arc from 60° to 300° going through 0°, which is the longer arc? Actually the shorter arc from 60° to 300° going through 0° is from 60° down to 0° then to -60° (300°)? Wait: 60° to 300° is 240° difference. The shorter arc is 120° (the other way). But we need the arc containing O(0,0) which is at angle 0°. So we need the arc that includes 0°. Starting at 60°, going clockwise to 300° passes through 0°? Let's see: 60° -> 0° (clockwise 60°), then 0° -> 300° (clockwise 60°). So total arc length central angle 120° (2π/3). Indeed the arc from 60° to 300° clockwise is 120°, not 240°. Because the minor arc from 60° to 300° is 120°? Actually angle difference: from 60° to 300°: if we go the shorter way, we can go from 60° down to 0° (60°), then to -60° (60°) total 120°. Yes. So arc length = (2π/3)*radius = 2π/3.

Thus region R total length (for P) is sum of two arcs = 4π/3.

But we need length of locus of Q, not P.

We need to compute length of curve Q = f(P) as P moves along R.

We can parametrize each arc with a parameter (angle) and compute the length integral.

Let P = (x,y) on R. Then AP = (x+2, y). |AP|^2 = (x+2)^2 + y^2.

Then Q = A + (9 / |AP|^2) * (P - A) = (-2,0) + (9 / ((x+2)^2 + y^2)) * (x+2, y).

So Q = ( -2 + 9(x+2)/D , 9y/D ) where D = (x+2)^2 + y^2.

Thus Q is a transformation of P.

We need to compute length of image curve. Possibly the transformation is an inversion-like transformation with center A and power 9? Indeed, if we consider inversion with center A and radius sqrt(9)=3, then inversion maps point P to Q' such that A, P, Q' collinear and AP * AQ' = 9. But here we have AP · AQ = 9 and they are parallel, but if they are in same direction, then AP * AQ = 9 (since dot product = |AP||AQ| cos0 = |AP||AQ|). Wait: AP and AQ are parallel, so they are either same direction or opposite direction. Dot product = |AP||AQ| cosθ where θ = 0 or π. If same direction, cosθ=1, so |AP||AQ| = 9. If opposite direction, cosθ=-1, so |AP||AQ| = -9. But dot product is given as 9, so it's positive, so same direction. Thus |AP||AQ| = 9. So indeed Q lies on ray from A through P such that AP * AQ = 9. That's exactly inversion in circle centered at A with radius 3 (since inversion: AP * AQ = r^2 = 9). So Q is the inverse of P with respect to circle centered at A with radius 3.

Thus mapping f is inversion: Q = inv_A, radius^2=9.

So we need length of image of curve R under inversion.

Inversion does not preserve lengths, but we can compute using parametrization.

We can parametrize each arc separately, compute Q(t), then compute length integral L = ∫ sqrt( (dx_Q/dt)^2 + (dy_Q/dt)^2 ) dt.

Alternatively, there is formula for length of inverse curve: If a curve is given in polar coordinates centered at A, inversion with radius a (a^2=9) maps radial distance r to r' = a^2 / r. The angle stays same. Then length element: ds' = (a^2 / r^2) ds? Not exactly: For a curve in polar coordinates (r,θ), arc length ds = sqrt( r^2 + (dr/dθ)^2 ) dθ. Under inversion r' = a^2 / r, we have dr'/dθ = -a^2 / r^2 dr/dθ. Then (ds')^2 = (r')^2 + (dr'/dθ)^2 dθ^2 = (a^4 / r^2) + (a^4 / r^4) (dr/dθ)^2 dθ^2 = a^4 / r^4 * ( r^2 + (dr/dθ)^2 ) dθ^2 = (a^4 / r^4) ds^2. So ds' = (a^2 / r^2) ds. But ds = original arc length element. So length of inverted curve = ∫ (a^2 / r^2) ds, where r is distance from A to point on curve.

Thus we can compute length of Q locus = ∫_R (9 / |AP|^2) ds_P, where ds_P is arc length element along R.

That's simpler: k = ∫_R (9 / ((x+2)^2 + y^2)) ds.

We need to compute this integral over the two arcs.

We can parametrize each arc.

Let’s do arc on C1: circle centered at O(0,0) radius 1. Parametrize with angle θ from 2π/3 to 4π/3? Actually we need the arc from R1 (θ=2π/3) to R2 (θ=4π/3) passing through π. So θ ∈ [2π/3, 4π/3].

Point P = (cosθ, sinθ). Distance from A(-2,0): |AP|^2 = (cosθ + 2)^2 + sin^2θ = cos^2θ + 4cosθ + 4 + sin^2θ = 1 + 4cosθ + 4 = 5 + 4cosθ.

Arc length element ds = 1 * dθ (since radius 1). So contribution from C1 arc: k1 = ∫_{θ=2π/3}^{4π/3} 9 / (5 + 4cosθ) dθ.

Arc on C2: circle centered at O'(-1,0) radius 1. Parametrize relative to center: P = (-1 + cosφ, sinφ). But we need arc from R1 to R2 containing O(0,0). R1 corresponds to φ = π/3? Let's compute: R1 = (-1/2, √3/2). Relative to center (-1,0): (-1/2 +1, √3/2) = (1/2, √3/2) => angle φ = π/3. R2: (1/2, -√3/2) => φ = -π/3 or 5π/3. O(0,0) relative to center: (1,0) => φ = 0. So the arc containing φ=0 between φ=π/3 and φ=-π/3 (or 5π/3) is the arc from φ=π/3 decreasing to 0 then to -π/3. So we can parametrize φ from π/3 to -π/3 (decreasing). Equivalent to φ ∈ [-π/3, π/3] traversed from π/3 to -π/3. We can integrate from φ = -π/3 to π/3 (increasing) and take same length.

Thus φ ∈ [-π/3, π/3].

Point P = (-1 + cosφ, sinφ). Distance from A(-2,0): |AP|^2 = (x+2)^2 + y^2 = (-1 + cosφ + 2)^2 + sin^2φ = (1 + cosφ)^2 + sin^2φ = 1 + 2cosφ + cos^2φ + sin^2φ = 2 + 2cosφ = 2(1+cosφ) = 4 cos^2(φ/2).

Arc length element ds = 1 * dφ.

Thus contribution from C2 arc: k2 = ∫_{φ=-π/3}^{π/3} 9 / (2(1+cosφ)) dφ = ∫ 9 / (4 cos^2(φ/2)) dφ? Wait: 2(1+cosφ) = 4 cos^2(φ/2). So integrand = 9 / (4 cos^2(φ/2)).

But we need to be careful: ds = dφ, not d(φ/2). So k2 = ∫_{-π/3}^{π/3} 9 / (4 cos^2(φ/2)) dφ.

Alternatively, we can compute numerically.

Total k = k1 + k2.

We need floor of k.

Let's compute using Python.

We'll compute integrals numerically with high precision.

[/THOUGHT] [PYTHON] import mpmath as mp

Define integrals

k1: integral from theta = 2π/3 to 4π/3 of 9/(5+4cosθ) dθ

def integrand1(theta): return 9/(5 + 4*mp.cos(theta))

k1 = mp.quad(integrand1, [2mp.pi/3, 4mp.pi/3]) print("k1 =", k1)

k2: integral from phi = -π/3 to π/3 of 9/(2*(1+cosφ)) dφ

def integrand2(phi): return 9/(2*(1 + mp.cos(phi)))

k2 = mp.quad(integrand2, [-mp.pi/3, mp.pi/3]) print("k2 =", k2)

k = k1 + k2 print("Total k =", k) print("Floor of k =", int(mp.floor(k))) [/PYTHON]

[THOUGHT] The computed total length (k \approx 17.762523), so (\lfloor k \rfloor = 17). [/THOUGHT] \boxed{17}