[THOUGHT] I'll use the fact that (Q) is the inversion of (P) with respect to the circle centered at (A(-2,0)) with radius (3): if (P) is at vector (\vec{v} = P-A), then [ Q = A + \frac{9}{|\vec{v}|^2},\vec{v}. ] I'll:

1. Compute intersection points (R_1,R_2) and key points (O',O,A).
2. Define the inversion map in Python.
3. Find images of three points on circle (C_1) (e.g., (R_1,R_2,O')) and determine the circle passing through their images (center and radius).
4. Determine which arc on this image-circle corresponds to the original arc (R_1O'R_2), compute its central angle and hence its arc length.
5. For circle (C_2) (which passes through the inversion center (A)), use the fact that its image is a straight line; confirm the images of (R_1,R_2) lie on that line and compute the segment length (|Q(R_1)Q(R_2)|).
6. Sum the arc length and segment length to get (k), then compute (\lfloor k\rfloor).

I'll use sympy for exact expressions (with radicals), then evaluate numerically to high precision and take the floor. [/THOUGHT] [PYTHON] import sympy as sp

Define symbols and points

x, y = sp.symbols('x y', real=True)

A = sp.Point(-2, 0)

Define inversion radius squared

R2 = 9 # radius^2 = 9, radius = 3

def invert_point(P): """Inversion of point P with respect to circle centered at A with radius^2 = R2.""" v = sp.Matrix([P.x - A.x, P.y - A.y]) r2 = v.dot(v) factor = sp.Rational(R2) / r2 v_new = factor * v Q = sp.Point(A.x + v_new[0], A.y + v_new[1]) return sp.simplify(Q)

Define circles C1: x^2 + y^2 = 1, C2: (x+1)^2 + y^2 = 1

Compute intersections R1, R2

solutions = sp.solve([ x2 + y2 - 1, (x+1)2 + y2 - 1 ], [x, y], dict=True)

solutions

Extract R1, R2

R1 = sp.Point(solutions[0][x], solutions[0][y]) R2 = sp.Point(solutions[1][x], solutions[1][y]) R1, R2

Define O', O

O_prime = sp.Point(-1, 0) O = sp.Point(0, 0)

Check they are on the intended circles

print("Check O' on C1:", sp.simplify(O_prime.x2 + O_prime.y2)) print("Check O on C2:", sp.simplify((O.x+1)2 + O.y2))

Inversion images for C1 arc points

Q_R1 = invert_point(R1) Q_R2 = invert_point(R2) Q_Oprime = invert_point(O_prime)

Q_R1, Q_R2, Q_Oprime

Find circle through Q_R1, Q_R2, Q_Oprime (image of C1)

circle_C1_image = sp.Circle(Q_R1, Q_R2, Q_Oprime) center_C1_img = circle_C1_image.center radius_C1_img = sp.simplify(circle_C1_image.radius)

center_C1_img, radius_C1_img

Compute central angles corresponding to arc Q(R1)->Q(O')->Q(R2)

We'll compute polar angles of these points relative to the image circle center.

def angle_at_center(C, P): """Angle of point P around center C in range (-pi, pi].""" return sp.atan2(P.y - C.y, P.x - C.x)

theta_R1 = sp.N(angle_at_center(center_C1_img, Q_R1)) theta_R2 = sp.N(angle_at_center(center_C1_img, Q_R2)) theta_Oprime = sp.N(angle_at_center(center_C1_img, Q_Oprime))

theta_R1, theta_Oprime, theta_R2

Determine central angle of the arc from R1 to R2 passing through O' image.

We want the smaller (or larger) angle that includes theta_Oprime.

We'll normalize to [0, 2*pi) and then choose correctly.

pi2 = 2*sp.pi

def normalize(angle): return sp.mod(angle + pi2, pi2)

nR1 = sp.N(normalize(theta_R1)) nR2 = sp.N(normalize(theta_R2)) nOprime = sp.N(normalize(theta_Oprime))

nR1, nOprime, nR2

Determine arc that goes from nR1 to nR2 passing through nOprime.

We'll consider two candidate arc lengths (in positive direction, and negative direction)

and pick the one where nOprime lies between nR1 and nR2 along that direction.

positive-direction angle from a to b on [0,2pi)

def pos_angle(a, b): return sp.mod(b - a + pi2, pi2)

arc_pos = sp.N(pos_angle(nR1, nR2)) arc_neg = pi2 - arc_pos # going the other way

arc_pos, arc_neg

Check if nOprime lies on positive-direction arc from nR1 to nR2

def is_on_pos_arc(a, b, c): """Check if c is on positive-direction arc from a to b.""" return 0 <= pos_angle(a, c) <= pos_angle(a, b) + 1e-12 # small tolerance

on_pos = is_on_pos_arc(float(nR1), float(nR2), float(nOprime)) on_pos

Similarly check for the other direction: from nR2 to nR1 (which corresponds to negative).

on_pos_reverse = is_on_pos_arc(float(nR2), float(nR1), float(nOprime)) on_pos_reverse

Determine correct central angle

if on_pos: central_angle_C1 = float(arc_pos) elif on_pos_reverse: central_angle_C1 = float(pos_angle(nR2, nR1)) else: # Fallback, but this should not happen central_angle_C1 = float(arc_pos)

central_angle_C1

Arc length of image of C1 arc

arc_length_C1 = float(radius_C1_img.evalf()) * central_angle_C1 arc_length_C1

Now handle C2: its image should be a line.

Invert R1, R2, O to see if they lie on a straight line.

Q_O = invert_point(O) Q_R1, Q_R2, Q_O

Verify collinearity

collinear = sp.Line(Q_R1, Q_R2).distance(Q_O).simplify() collinear

Length of image of arc R1OR2 on C2 is just distance between Q_R1 and Q_R2

segment_length_C2 = float(Q_R1.distance(Q_R2).evalf()) segment_length_C2

Total length k

k = arc_length_C1 + segment_length_C2 k, float(k)

Floor of k

import math math.floor(k)

[THOUGHT] From the inversion geometry, the locus of (Q) corresponding to the arc of (C_1) is an arc of a circle of radius (3) with central angle (4\pi/3), giving length (4\pi). The locus corresponding to the arc of (C_2) is a line segment of length (3\sqrt{3}). Thus [ k = 4\pi + 3\sqrt{3} \approx 17.76, ] so (\lfloor k \rfloor = 17). [/THOUGHT] \boxed{17}