[THOUGHT] We need to understand the problem fully.

Given A(-2,0). Points P and Q such that: (i) Vectors AP and AQ are parallel. That means Q lies on the line through A and P (same line, possibly opposite direction). So Q = A + t*(P - A) for some scalar t (real). (ii) Dot product AP · AQ = 9.

Let v = P - A = (x+2, y) where P = (x,y). Then AP = v, AQ = t v. Dot product: v · (t v) = t |v|^2 = 9 ⇒ t = 9 / |v|^2.

Thus Q = A + (9/|v|^2) v.

So given P, Q is determined uniquely (except if v=0, but P ≠ A? Possibly P can be A? But then v=0, t undefined, but dot product would be 0, not 9, so P cannot be A. So v ≠ 0.)

Now region R consists of two arcs:

• Arc R1 O' R2 of circle C1: x^2 + y^2 = 1, containing point O'(-1,0). So arc from R1 to R2 passing through O'(-1,0). That arc is the left half of C1? Actually O' is on C1: (-1,0) satisfies x^2+y^2=1. So arc containing O' is the left semicircle? Wait: C1 is unit circle centered at origin. Points R1 and R2 are intersections of C1 and C2. Let's find them.

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

Intersection: subtract equations: (x+1)^2 + y^2 - (x^2 + y^2) = 1 - 1 ⇒ (x^2+2x+1) - x^2 = 0 ⇒ 2x+1=0 ⇒ x = -1/2.

Then from C1: x^2 + y^2 = 1 ⇒ 1/4 + y^2 = 1 ⇒ y^2 = 3/4 ⇒ y = ±√3/2.

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

Arc R1 O' R2 of C1 containing O'(-1,0): The arc from C1 connecting R1 to R2 passing through O'. That arc is the left part of the unit circle from R1 to R2 going through O'. Since O' is at (-1,0) which is leftmost point, the arc is the left semicircle? Actually the left semicircle would be from R1 to R2 via O'? Let's check: The left semicircle x ≤ 0. R1 and R2 have x = -1/2 < 0, O' has x = -1. So yes, the arc from R1 to R2 passing through O' is the left semicircle (including the point at leftmost). But note: The arc containing O' is the longer arc? Wait: There are two arcs between R1 and R2 on C1: one is the left arc (passing through O') of length π, the other is the right arc (passing through (1,0)) of length π. Since O' is on the left arc, the specified arc is the left semicircle.

Similarly, arc R1 O R2 of C2 containing O(0,0). C2 is circle centered at (-1,0) radius 1. O = (0,0) lies on C2? Check: (0+1)^2 + 0^2 = 1, yes O is on C2. So arc from R1 to R2 passing through O is the arc of C2 containing O. That arc is also length π? Actually C2's center is (-1,0). Points R1 and R2 are at (-1/2, ±√3/2). O is at (0,0). The arc containing O is the right semicircle of C2 (since O is rightmost point on C2? Actually C2's rightmost point is (0,0) because center (-1,0) radius 1, so rightmost is (0,0). So O is the rightmost point. So the arc from R1 to R2 passing through O is the right semicircle of C2.

Thus region R is the union of two semicircular arcs: left semicircle of C1 and right semicircle of C2. They share the chord R1R2. So region R is like a "lens" shape but consisting of two semicircles? Actually it's the boundary of a shape? Wait: The problem says "region R consisting of the arc R1 O' R2 of circle C1, which contains the point O'(-1,0), together with the arc R1 O R2 of circle C2, which contains the point O(0,0)." So region R is the closed curve formed by these two arcs. So R is the boundary of a region shaped like a lens but with two different circles. Actually the two arcs together form a closed curve: starting from R1, go along C1 arc to O' to R2, then back along C2 arc from R2 to O to R1. That's a closed curve. So region R is that closed curve (the boundary). But the problem says "When the point P moves along the region R" — likely meaning P moves along this closed curve (the boundary). So P is on that curve.

Now we need the locus traced by Q as P moves along this curve. Then k is total length of that locus (the curve traced by Q). Then we need floor(k).

So we need to find the set of Q for P on the closed curve R.

Given Q = A + (9/|v|^2) v, where v = P - A, A = (-2,0). So v = (x+2, y). |v|^2 = (x+2)^2 + y^2.

Thus Q = (-2,0) + 9 * (x+2, y) / ((x+2)^2 + y^2).

Let’s denote u = (X,Y) = Q. Then:

X = -2 + 9*(x+2) / ((x+2)^2 + y^2) Y = 0 + 9*y / ((x+2)^2 + y^2)

We have P = (x,y) lies on the union of two arcs: arc1: x^2 + y^2 = 1, with x ≤ -1/2? Actually arc1 is left semicircle of C1: x^2 + y^2 = 1 and x ≤ -1/2? Wait: The arc from R1 to R2 via O' includes all points on unit circle with x ≤ -1/2? Not exactly: The left semicircle x ≤ 0 includes all points with x ≤ 0. But our arc is specifically the left semicircle from R1 to R2 passing through O'. That includes points with x between -1 and -1/2? Actually O' is at (-1,0). R1 and R2 have x = -1/2. So the arc is the set of points on unit circle with x ∈ [-1, -1/2] and y such that y^2 = 1 - x^2. But careful: The arc from R1 to R2 via O' goes along the unit circle in the left half-plane, but does it include points with x between -1 and -1/2? Yes, because moving from R1 (x=-1/2) to O' (x=-1) along the unit circle, x decreases from -1/2 to -1, then increases back to -1/2 at R2. So the arc is the set of points on unit circle with x ∈ [-1, -1/2] (but note for each x there are two y values symmetric, except at x=-1). However, the arc includes both upper and lower halves? Actually the arc from R1 to R2 via O' includes both the upper and lower halves? Wait: The arc is a continuous curve from R1 to R2 passing through O'. Starting at R1 (-1/2, √3/2), going along the unit circle in the direction of decreasing x? There are two possible arcs from R1 to R2: one going through O' (left arc) and one going through the right side. The left arc passes through O' and goes through the lower half? Let's param: For unit circle, angle θ from positive x-axis. R1 corresponds to θ = 2π/3 (120°), R2 corresponds to θ = 4π/3 (240°). O' corresponds to θ = π (180°). The left arc from R1 to R2 via O' goes from θ = 2π/3 down to θ = π then to θ = 4π/3. That's the arc with θ ∈ [2π/3, 4π/3]. That includes both upper and lower halves? Actually θ from 2π/3 to π is upper half (positive y), θ from π to 4π/3 is lower half (negative y). So indeed the arc includes both upper and lower halves, symmetric.

Similarly, arc2: C2: (x+1)^2 + y^2 = 1, containing O(0,0). O corresponds to θ = 0 in polar coordinates centered at (-1,0)? Let's param: For C2, center (-1,0), radius 1. Let angle φ measured from positive x-direction relative to center. Then points: (x,y) = (-1 + cos φ, sin φ). O = (0,0) corresponds to cos φ = 1, sin φ = 0 ⇒ φ = 0. R1 = (-1/2, √3/2): x = -1/2 = -1 + cos φ ⇒ cos φ = 1/2, y = √3/2 = sin φ ⇒ φ = π/3. R2 = (-1/2, -√3/2): cos φ = 1/2, sin φ = -√3/2 ⇒ φ = -π/3 (or 5π/3). The arc containing O(0,0) from R1 to R2: which arc? There are two arcs: one going from φ = π/3 to φ = -π/3 passing through φ = 0 (that's the right arc), and the other going the long way around. The arc containing φ = 0 is the right semicircle, with φ ∈ [-π/3, π/3]. So arc2 is points on C2 with φ ∈ [-π/3, π/3], i.e., x = -1 + cos φ, y = sin φ, with φ from -π/3 to π/3.

Thus P is on union of two arcs: Arc1: unit circle, θ ∈ [2π/3, 4π/3]; Arc2: circle C2, φ ∈ [-π/3, π/3].

Now we need the locus of Q as P moves along these arcs.

We can attempt to derive the transformation from P to Q.

Let’s denote P = (x,y). Let D = |AP|^2 = (x+2)^2 + y^2.

Then Q = (X,Y) = (-2 + 9*(x+2)/D, 9*y/D).

We can invert: Given Q, what is P? Let’s solve for x,y in terms of X,Y.

From Q = A + 9 v / |v|^2, where v = P - A.

Thus Q - A = 9 v / |v|^2. Let w = Q - A = (X+2, Y). Then w = 9 v / |v|^2.

Thus v = (9 / |w|^2) w? Wait: w = 9 v / |v|^2 ⇒ multiply both sides by |v|^2: w |v|^2 = 9 v.

But also v = (9/|w|^2) w? Let's derive: Since w = 9 v / |v|^2, then |w| = 9 / |v| because |v|^2 in denominator gives factor 1/|v| times 9/|v|? Actually |w| = |9 v / |v|^2| = 9/|v|. So |v| = 9/|w|.

Also direction: v is parallel to w (since w is scalar multiple of v). So v = λ w for some λ >0? Actually w = (9/|v|^2) v, so λ = 9/|v|^2. But we have |v| = 9/|w|, so λ = 9/( (81/|w|^2) ) = (|w|^2)/9.

Thus v = (|w|^2 / 9) w.

Check: v = (|w|^2 / 9) w. Then |v| = (|w|^2 / 9) |w| = |w|^3 / 9. But earlier we had |v| = 9/|w|. Equate: |w|^3 / 9 = 9/|w| ⇒ |w|^4 = 81 ⇒ |w| = √3? That seems contradictory unless |w| is fixed. Wait, I made a mistake.

Let's derive carefully:

Given w = Q - A = 9 v / |v|^2.

Then w is parallel to v, so v = μ w for some scalar μ. Substitute: w = 9 (μ w) / |μ w|^2 = 9 μ w / (μ^2 |w|^2) = (9/μ) w / |w|^2.

Thus w = (9/(μ |w|^2)) w ⇒ 1 = 9/(μ |w|^2) ⇒ μ = 9/|w|^2.

Thus v = μ w = (9/|w|^2) w.

Now |v| = |μ w| = μ |w| = (9/|w|^2) |w| = 9/|w|. So indeed |v| = 9/|w|.

Also, from earlier, we had |w| = 9/|v|, which matches.

Thus the transformation is involutive: P = A + v, with v = (9/|w|^2) w, where w = Q - A.

So given Q, we can find P. But we need the locus of Q when P lies on the arcs.

Alternatively, we can param P on each arc and compute Q.

Let's param:

Arc1: P(θ) = (cos θ, sin θ), θ ∈ [2π/3, 4π/3].

Then v = (cos θ + 2, sin θ). D = (cos θ + 2)^2 + sin^2 θ = cos^2 θ + 4 cos θ + 4 + sin^2 θ = 1 + 4 cos θ + 4 = 5 + 4 cos θ.

Thus Q(θ) = (-2 + 9*(cos θ + 2)/D, 9*sin θ/D).

Simplify: D = 5 + 4 cos θ.

So Q(θ) = ( -2 + 9(cos θ + 2)/(5 + 4 cos θ), 9 sin θ/(5 + 4 cos θ) ).

Arc2: P(φ) = (-1 + cos φ, sin φ), φ ∈ [-π/3, π/3].

Then v = (x+2, y) = (-1 + cos φ + 2, sin φ) = (1 + cos φ, sin φ).

D = (1 + cos φ)^2 + sin^2 φ = 1 + 2 cos φ + cos^2 φ + sin^2 φ = 2 + 2 cos φ = 2(1 + cos φ) = 4 cos^2(φ/2).

Thus Q(φ) = (-2 + 9*(1 + cos φ)/D, 9*sin φ/D).

Compute: D = 2(1 + cos φ) = 4 cos^2(φ/2). Also 1+cos φ = 2 cos^2(φ/2).

So 9*(1+cos φ)/D = 9 * (2 cos^2(φ/2)) / (4 cos^2(φ/2)) = 9/2.

Thus X = -2 + 9/2 = 5/2.

And Y = 9 sin φ / D = 9 sin φ / (2(1+cos φ)) = (9/2) * sin φ/(1+cos φ) = (9/2) * tan(φ/2) (since sin φ/(1+cos φ) = tan(φ/2)).

Thus Q(φ) = (5/2, (9/2) tan(φ/2) ).

As φ ranges from -π/3 to π/3, φ/2 ranges from -π/6 to π/6, tan(φ/2) ranges from tan(-π/6) = -1/√3 to tan(π/6) = 1/√3.

Thus on arc2, Q moves along the vertical line X = 5/2, Y from -9/(2√3) to 9/(2√3). So it's a straight segment.

Now for arc1, we have Q(θ) param equation. That's more complicated. We need to see if that traces some curve. Possibly a circle or ellipse.

Let's compute Q(θ) explicitly.

Let’s denote D = 5 + 4 cos θ.

Then X = -2 + 9(cos θ + 2)/D = (-2D + 9(cos θ + 2))/D = (-2(5+4 cos θ) + 9 cos θ + 18)/D = (-10 -8 cos θ + 9 cos θ + 18)/D = (8 + cos θ)/D.

So X = (8 + cos θ) / (5 + 4 cos θ).

Y = 9 sin θ / D = 9 sin θ / (5 + 4 cos θ).

We can try to eliminate θ to find relation between X and Y.

Let’s set t = tan(θ/2). Then cos θ = (1 - t^2)/(1 + t^2), sin θ = 2t/(1 + t^2).

But maybe easier: Compute X^2 + Y^2? Or find Cartesian equation.

We have X = (8 + cos θ)/(5 + 4 cos θ), Y = 9 sin θ/(5 + 4 cos θ).

Let’s denote u = cos θ, v = sin θ, with u^2 + v^2 = 1.

Then X = (8 + u)/(5 + 4u), Y = 9 v/(5 + 4u).

We can solve for u in terms of X: X(5 + 4u) = 8 + u ⇒ 5X + 4X u = 8 + u ⇒ 4X u - u = 8 - 5X ⇒ u(4X - 1) = 8 - 5X ⇒ u = (8 - 5X)/(4X - 1).

Similarly, from Y: Y(5 + 4u) = 9 v ⇒ v = Y(5 + 4u)/9.

But v^2 = 1 - u^2.

Thus (Y(5 + 4u)/9)^2 = 1 - u^2.

Substitute u expression in terms of X.

This will give an equation relating X and Y. However, note that P is on unit circle, so u^2 + v^2 = 1. But also u = cos θ, v = sin θ, with θ ∈ [2π/3, 4π/3]. That corresponds to u ∈ [-1, -1/2] and v can be positive or negative.

But we can compute the range of Q on arc1. Possibly the locus is also a circular arc or line segment.

Let's compute numerically for a few θ values to see the shape.

We'll write Python to compute Q for θ in that range and maybe see if it's an arc of a circle.

But first, note that arc2 gives a vertical segment. Arc1 likely gives another arc. The total locus of Q as P moves along the closed curve R (union of arc1 and arc2) will be the union of the image of arc1 and the image of arc2 under the transformation.

Thus k = length of image of arc1 + length of image of arc2.

We already have image of arc2: vertical line segment from Y = -9/(2√3) to Y = 9/(2√3), length = 9/√3 = 3√3 ≈ 5.19615.

Now image of arc1: we need to compute the curve traced by Q(θ) as θ goes from 2π/3 to 4π/3. That's a continuous curve. We need its length.

We can compute length via integral: L1 = ∫_{θ=2π/3}^{4π/3} sqrt( (dX/dθ)^2 + (dY/dθ)^2 ) dθ.

We can compute numerically.

But maybe the curve is a circle or line segment as well? Let's check numerically.

We'll compute Q for several θ values and see if they lie on a circle or line.

Let's write Python to compute Q and also compute distance from some center.

But before that, note that arc1 is the left semicircle of unit circle. Possibly the transformation maps circles through A to lines? Because Q = A + 9 v/|v|^2 is like an inversion with center A and radius √9? Actually inversion: Inversion with center A and radius r maps point P (≠ A) to Q such that Q lies on ray AP and |AP| * |AQ| = r^2. Here we have Q = A + 9 v/|v|^2, so |AQ| = 9/|v|, and |AP| = |v|, so |AP| * |AQ| = 9. Indeed, that's inversion with radius √9 = 3. So Q is the inversion of P with respect to a circle centered at A with radius 3.

Thus the transformation is inversion: P → Q such that A, P, Q collinear and AP * AQ = 9.

Thus the locus of Q is the inverse of the curve R under inversion centered at A(-2,0) with radius 3.

So we need to find the inverse of the closed curve R (union of two arcs) under this inversion.

Inversion properties: A circle through A inverts to a line not through A. A circle not through A inverts to another circle. A line through A inverts to itself.

Here R consists of two arcs: arc1 is part of unit circle C1: x^2 + y^2 = 1. Does this circle pass through A(-2,0)? Check: (-2)^2 + 0^2 = 4 ≠ 1, so C1 does not pass through A. So its inverse will be another circle.

Arc2 is part of circle C2: (x+1)^2 + y^2 = 1. Does C2 pass through A? Plug A: (-2+1)^2 + 0^2 = 1, yes! Because (-1)^2 = 1. So C2 passes through A. Therefore, the arc of C2 (which is part of C2) inverts to a line (since circle through A inverts to line).

Indeed, we earlier computed that for arc2, Q lies on vertical line X = 5/2. That matches: the image of C2 (the whole circle) under inversion is a line. Since our arc2 is a subset of C2, its image is a subset of that line. We found it's a segment of that line.

Thus the locus of Q is the union of:

• The image of arc1 (part of C1 not through A) which is an arc of a circle (the inverse of C1).
• The image of arc2 (part of C2 through A) which is a line segment (as computed).

Now we need to find the exact circle that is the inverse of C1.

C1: x^2 + y^2 = 1. Inversion centered at A(-2,0) with radius r=3.

General inversion: For point (x,y), its inverse (X,Y) satisfies: (X - x_A, Y - y_A) = r^2 (x - x_A, y - y_A) / |(x - x_A, y - y_A)|^2.

Here A = (-2,0), r=3.

So (X+2, Y) = 9 (x+2, y) / ((x+2)^2 + y^2). Which matches earlier.

Now we want the image of circle C1 under this inversion. Since C1 does not pass through A, its image is a circle.

We can find the equation of that circle.

Let’s denote u = x+2, v = y. Then (x,y) = (u-2, v). C1 equation: (u-2)^2 + v^2 = 1 ⇒ u^2 -4u +4 + v^2 =1 ⇒ u^2 + v^2 -4u +3 =0.

Inversion: (X+2, Y) = 9 (u, v) / (u^2 + v^2).

Let w = X+2, z = Y. Then w = 9u/(u^2+v^2), z = 9v/(u^2+v^2).

We want to find the locus of (w,z) as (u,v) satisfies u^2+v^2 -4u +3=0.

We can express u,v in terms of w,z: u = (9/(w^2+z^2)) w, v = (9/(w^2+z^2)) z? Wait: from inversion formula: (w,z) = 9 (u,v)/(u^2+v^2). So (u,v) = (w^2+z^2) (w,z)/9.

Thus u = (w^2+z^2) w / 9, v = (w^2+z^2) z / 9.

Now plug into C1 equation in u,v: u^2+v^2 -4u +3=0.

Compute u^2+v^2 = ((w^2+z^2)^2 /81) (w^2+z^2) = (w^2+z^2)^3 /81.

And -4u = -4 (w^2+z^2) w /9.

Thus equation: (w^2+z^2)^3 /81 - 4 (w^2+z^2) w /9 + 3 = 0.

Multiply by 81: (w^2+z^2)^3 - 36 (w^2+z^2) w + 243 = 0.

That's the equation of the inverse circle in w,z coordinates. But w = X+2, z = Y. So in terms of X,Y: ((X+2)^2 + Y^2)^3 - 36 ((X+2)^2 + Y^2)(X+2) + 243 = 0.

That's a circle? Not obviously a circle. However, inversion of a circle not through A is a circle. So the equation should represent a circle. Let's try to simplify.

Let ρ^2 = (X+2)^2 + Y^2 = w^2+z^2. Then equation: ρ^6 - 36 ρ^2 w + 243 = 0, where w = X+2.

So ρ^6 - 36 w ρ^2 + 243 = 0.

This is symmetric in Y, so likely a circle centered on the X-axis.

Let’s attempt to find center and radius. Write ρ^2 = (X+2)^2 + Y^2 = X^2 + 4X +4 + Y^2.

But maybe we can find explicit circle equation by plugging a few points.

Alternatively, we can find the image of C1 by taking three points on C1, inverting them, and finding the circle through those three points.

Points on C1: we can take R1, R2, and O'(-1,0) maybe. But careful: R1 and R2 are on C1, but are they on the arc1? Yes, arc1 includes R1 and R2. O' is also on arc1. So these three points map to three points on image circle.

Compute Q for these points.

First, O' = (-1,0). v = O' - A = (-1+2, 0) = (1,0). |v|^2 = 1. So Q = A + 9 v / |v|^2 = (-2,0) + 9*(1,0) = (7,0).

Second, R1 = (-1/2, √3/2). v = (-1/2+2, √3/2) = (3/2, √3/2). |v|^2 = (9/4) + (3/4) = 12/4 = 3. So Q = (-2,0) + 9*(3/2, √3/2)/3 = (-2,0) + (9/3)(3/2, √3/2) = (-2,0) + 3(3/2, √3/2) = (-2,0) + (9/2, 3√3/2) = (9/2 - 2, 3√3/2) = (5/2, 3√3/2).

Third, R2 = (-1/2, -√3/2). v = (3/2, -√3/2). |v|^2 = 3. Q = (-2,0) + 9*(3/2, -√3/2)/3 = (-2,0) + 3*(3/2, -√3/2) = (-2,0) + (9/2, -3√3/2) = (5/2, -3√3/2).

So we have three points on the image circle: A' = (7,0), B' = (5/2, 3√3/2), C' = (5/2, -3√3/2).

These three points: A' is symmetric about x-axis, B' and C' symmetric. So the circle is centered on x-axis. Let center be (h,0). Radius R.

Distance from center to A': |h - 7|. Distance to B': sqrt((h - 5/2)^2 + (3√3/2)^2) = sqrt((h - 5/2)^2 + 27/4).

Set equal: (h-7)^2 = (h - 5/2)^2 + 27/4.

Expand: h^2 -14h +49 = h^2 -5h + 25/4 + 27/4 = h^2 -5h + 52/4 = h^2 -5h +13.

Cancel h^2: -14h +49 = -5h +13 ⇒ -9h = -36 ⇒ h = 4.

Thus center is (4,0). Radius = distance from center to A' = |4-7| = 3. Or distance to B': sqrt((4 - 5/2)^2 + 27/4) = sqrt((3/2)^2 + 27/4) = sqrt(9/4 + 27/4) = sqrt(36/4) = sqrt(9) = 3. Yes.

So the image of entire C1 under inversion is circle centered at (4,0) radius 3.

But our arc1 is only the part from R1 to R2 via O'. So the image arc is part of this circle. Which part? We need to see which portion of this circle corresponds to the arc1.

We already have images of O', R1, R2 on the circle: O' maps to (7,0), R1 maps to (5/2, 3√3/2), R2 maps to (5/2, -3√3/2). So the arc on C1 (left semicircle) maps to the arc of this circle between images of R1 and R2 passing through image of O'. Let's see the circle center (4,0) radius 3. Points: (7,0) is rightmost point. (5/2, 3√3/2) is at angle? Compute angle from center: vector (5/2 - 4, 3√3/2) = (-3/2, 3√3/2). That's length sqrt(9/4 + 27/4) = sqrt(36/4)=3. So angle = arctan((3√3/2)/(-3/2)) = arctan(-√3) = -π/3? Actually x negative, y positive => second quadrant, angle = π - π/3 = 2π/3. Similarly, (5/2, -3√3/2) gives angle = -2π/3 (or 4π/3). So the arc from R1 to R2 via O' on C1 corresponds to arc on the image circle from angle 2π/3 to angle -2π/3 passing through angle 0? Wait: O' maps to (7,0) which is angle 0. So the arc on C1 goes from R1 (θ=2π/3) to O' (θ=π) to R2 (θ=4π/3). The image points: R1 maps to angle 2π/3, O' maps to angle 0, R2 maps to angle -2π/3 (or 4π/3). So the image arc goes from angle 2π/3 down to 0 then to -2π/3. That's the left side of the circle? Actually angles measured from positive x-axis: 2π/3 is 120°, 0 is 0°, -2π/3 is -120°. So the arc spans angles from -2π/3 to 2π/3 passing through 0. That's the right half of the circle? Wait: center (4,0). The circle radius 3. The point (7,0) is rightmost. Points at angles ±2π/3 are left side? Actually from center (4,0), the point at angle 2π/3 has coordinates (4+3cos(2π/3), 3sin(2π/3)) = (4+3*(-1/2), 3*(√3/2)) = (4 - 1.5, (3√3/2)) = (2.5, 3√3/2) which matches (5/2, 3√3/2). Yes. So that's the left side? Actually x=2.5, center at x=4, so left of center. So the arc from angle 2π/3 to -2π/3 going through 0 is the arc that goes through the rightmost point? Wait: angle 0 gives (7,0). So the arc from angle 2π/3 to -2π/3 passing through 0 goes from left side (120°) down to right side (0°) then down to left side (-120°). That's the arc that covers the rightmost point and goes around through top and bottom? Actually the circle: angles from -2π/3 to 2π/3 is the arc that includes the rightmost point (angle 0) and excludes the leftmost point (angle π). So indeed it's the right half of the circle? But note the circle center (4,0) radius 3, leftmost point is (1,0). That's not included because angle π gives x=4+3cosπ=1. So our arc is the right half of the circle (the half with angles between -2π/3 and 2π/3). That's an arc of length: radius * angle difference = 3 * (2π/3 - (-2π/3)) = 3 * (4π/3) = 4π ≈ 12.566.

But wait: The arc on C1 is length π (since it's a semicircle). Under inversion, length is not preserved. So the image arc length is not necessarily π. We need to compute the length of the image arc of arc1.

Thus the image of arc1 is an arc of the circle centered at (4,0) radius 3, spanning angles from -2π/3 to 2π/3 (or equivalently from 2π/3 to 4π/3? Actually we need to be careful: The mapping from C1 to this circle is bijective? Inversion maps the entire circle C1 (except point A) to the circle we found. Since C1 does not pass through A, the inversion is a bijection between C1 and the image circle. So the arc on C1 maps to an arc on the image circle. The arc on C1 is from R1 to R2 via O'. That arc corresponds to angles on C1 from 2π/3 to 4π/3. Under inversion, each point on C1 maps to a point on the image circle. The mapping is not linear in angle, but we can compute the image arc length directly via the integral formula.

Thus we need to compute length of the image arc of arc1. We can compute using parametric equations Q(θ) and integrate length.

Similarly, for arc2, we already have length = 3√3.

Thus total k = length of image arc1 + length of image arc2.

We'll compute numerically.

Let's write Python to compute length1 via numerical integration of sqrt(dX/dθ^2 + dY/dθ^2) over θ from 2π/3 to 4π/3.

We'll also verify that the image arc indeed lies on that circle.

Then compute k and floor(k).

We'll use high precision.

Let's code. </think>

[PYTHON] import numpy as np import math

Define transformation for arc1

def Q_arc1(theta): # theta in radians cos_t = math.cos(theta) sin_t = math.sin(theta) D = 5 + 4*cos_t X = (8 + cos_t) / D Y = 9 * sin_t / D return X, Y

Derivatives dX/dtheta, dY/dtheta

def dQ_arc1(theta, h=1e-8): X1, Y1 = Q_arc1(theta + h) X0, Y0 = Q_arc1(theta - h) dX = (X1 - X0) / (2h) dY = (Y1 - Y0) / (2h) return dX, dY

Compute length of arc1 via numerical integration

theta_start = 2math.pi/3 theta_end = 4math.pi/3 n = 100000 dtheta = (theta_end - theta_start) / n length1 = 0.0 for i in range(n): theta = theta_start + (i + 0.5) * dtheta # midpoint dX, dY = dQ_arc1(theta) length1 += math.sqrt(dXdX + dYdY) * dtheta

print("Length of image arc1 (arc1 of C1):", length1)

Length of image arc2 (vertical segment)

length2 = 9 / math.sqrt(3) # 3*sqrt(3) print("Length of image arc2 (arc2 of C2):", length2)

k = length1 + length2 print("Total k:", k) print("Floor(k):", math.floor(k)) [/PYTHON]

[THOUGHT] The total length (k) is approximately (17.7625), so (\lfloor k \rfloor = 17).

\boxed{17}