[THOUGHT] I need to solve the problem step by step.

First, part (1): Compute the area enclosed by the curve ( r = 2 + \cos\theta ) for ( 0 \le \theta \le 2\pi ).

In polar coordinates, the area enclosed by a curve ( r(\theta) ) from ( \theta = \alpha ) to ( \theta = \beta ) is given by: [ A = \frac{1}{2} \int_{\alpha}^{\beta} r(\theta)^2 , d\theta. ] Here, ( r(\theta) = 2 + \cos\theta ), and the curve is traced once as ( \theta ) goes from 0 to ( 2\pi ). So: [ A = \frac{1}{2} \int_{0}^{2\pi} (2 + \cos\theta)^2 , d\theta. ] Expand: ( (2 + \cos\theta)^2 = 4 + 4\cos\theta + \cos^2\theta ).

Thus: [ A = \frac{1}{2} \int_{0}^{2\pi} \left(4 + 4\cos\theta + \cos^2\theta\right) d\theta. ] We can compute each integral separately.

Note: ( \int_{0}^{2\pi} \cos\theta , d\theta = 0 ).

Also, ( \cos^2\theta = \frac{1 + \cos 2\theta}{2} ).

So: [ \int_{0}^{2\pi} \cos^2\theta , d\theta = \int_{0}^{2\pi} \frac{1 + \cos 2\theta}{2} , d\theta = \frac{1}{2} \int_{0}^{2\pi} 1 , d\theta + \frac{1}{2} \int_{0}^{2\pi} \cos 2\theta , d\theta. ] The second integral is zero because ( \cos 2\theta ) over a full period (actually two periods) integrates to zero. So: [ \int_{0}^{2\pi} \cos^2\theta , d\theta = \frac{1}{2} \cdot 2\pi = \pi. ]

Now: [ \int_{0}^{2\pi} 4 , d\theta = 4 \cdot 2\pi = 8\pi. ] [ \int_{0}^{2\pi} 4\cos\theta , d\theta = 0. ] So: [ \int_{0}^{2\pi} (4 + 4\cos\theta + \cos^2\theta) , d\theta = 8\pi + 0 + \pi = 9\pi. ] Thus: [ A = \frac{1}{2} \cdot 9\pi = \frac{9\pi}{2}. ] So ( a = \frac{9\pi}{2} ).

Now part (2): At the point ( (r, \theta) = \left( \frac{4+\sqrt{2}}{2}, \frac{\pi}{4} \right) ), find the slope of the tangent line to the curve ( C ) in the ( xy )-plane.

We have ( r = 2 + \cos\theta ). Check if the given point satisfies this: at ( \theta = \pi/4 ), ( \cos(\pi/4) = \sqrt{2}/2 ), so: [ r = 2 + \frac{\sqrt{2}}{2} = \frac{4 + \sqrt{2}}{2}. ] Yes, it matches.

We need the slope ( \frac{dy}{dx} ) at that point. In polar coordinates: [ x = r\cos\theta, \quad y = r\sin\theta. ] Then: [ \frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}}. ] We have: [ \frac{dy}{d\theta} = \frac{dr}{d\theta} \sin\theta + r \cos\theta, ] [ \frac{dx}{d\theta} = \frac{dr}{d\theta} \cos\theta - r \sin\theta. ] Here, ( r = 2 + \cos\theta ), so ( \frac{dr}{d\theta} = -\sin\theta ).

Thus: [ \frac{dy}{d\theta} = (-\sin\theta) \sin\theta + r \cos\theta = -\sin^2\theta + r \cos\theta, ] [ \frac{dx}{d\theta} = (-\sin\theta) \cos\theta - r \sin\theta = -\sin\theta \cos\theta - r \sin\theta. ]

At ( \theta = \pi/4 ): [ \sin(\pi/4) = \cos(\pi/4) = \frac{\sqrt{2}}{2}. ] So: [ \sin^2(\pi/4) = \left( \frac{\sqrt{2}}{2} \right)^2 = \frac{2}{4} = \frac{1}{2}. ] [ r = \frac{4 + \sqrt{2}}{2}. ] Now compute: [ \frac{dy}{d\theta} = -\frac{1}{2} + r \cdot \frac{\sqrt{2}}{2} = -\frac{1}{2} + \frac{4 + \sqrt{2}}{2} \cdot \frac{\sqrt{2}}{2}. ] Simplify: [ \frac{4 + \sqrt{2}}{2} \cdot \frac{\sqrt{2}}{2} = \frac{(4 + \sqrt{2})\sqrt{2}}{4} = \frac{4\sqrt{2} + 2}{4} = \frac{2\sqrt{2} + 1}{2}. ] Wait, let's compute carefully: [ (4 + \sqrt{2}) \cdot \sqrt{2} = 4\sqrt{2} + 2. ] Divide by 4: ( \frac{4\sqrt{2} + 2}{4} = \frac{2\sqrt{2} + 1}{2} ). So: [ \frac{dy}{d\theta} = -\frac{1}{2} + \frac{2\sqrt{2} + 1}{2} = \frac{-\frac{1}{2} \cdot 2? Wait: better common denominator 2:} ] [ -\frac{1}{2} + \frac{2\sqrt{2} + 1}{2} = \frac{-1 + 2\sqrt{2} + 1}{2} = \frac{2\sqrt{2}}{2} = \sqrt{2}. ] So ( \frac{dy}{d\theta} = \sqrt{2} ).

Now ( \frac{dx}{d\theta} = -\sin\theta \cos\theta - r \sin\theta ). At ( \theta = \pi/4 ): [ \sin\theta \cos\theta = \left( \frac{\sqrt{2}}{2} \right)^2 = \frac{1}{2}. ] So: [ \frac{dx}{d\theta} = -\frac{1}{2} - r \cdot \frac{\sqrt{2}}{2}. ] We already computed ( r \cdot \frac{\sqrt{2}}{2} = \frac{2\sqrt{2} + 1}{2} ). Thus: [ \frac{dx}{d\theta} = -\frac{1}{2} - \frac{2\sqrt{2} + 1}{2} = \frac{-1 - 2\sqrt{2} - 1}{2} = \frac{-2 - 2\sqrt{2}}{2} = -1 - \sqrt{2}. ]

Therefore: [ \frac{dy}{dx} = \frac{\sqrt{2}}{-1 - \sqrt{2}}. ] Simplify by multiplying numerator and denominator by ( -1 + \sqrt{2} ): [ \frac{\sqrt{2}}{-1 - \sqrt{2}} \cdot \frac{-1 + \sqrt{2}}{-1 + \sqrt{2}} = \frac{\sqrt{2}(-1 + \sqrt{2})}{(-1)^2 - (\sqrt{2})^2} = \frac{-\sqrt{2} + 2}{1 - 2} = \frac{2 - \sqrt{2}}{-1} = -2 + \sqrt{2}. ] So the slope is ( -2 + \sqrt{2} ).

The problem asks: "Express the answer from part (2) in the form ( b + \sqrt{c} ), where ( b ) and ( c ) are integers." Here, slope = ( -2 + \sqrt{2} ), so ( b = -2 ), ( c = 2 ).

But wait: the problem says "let those integers be the definitions of ( b ) and ( c )." So ( b = -2 ), ( c = 2 ).

Now we need to compute ( \frac{abc}{\pi} ), where ( a = \frac{9\pi}{2} ), ( b = -2 ), ( c = 2 ).

Thus: [ \frac{abc}{\pi} = \frac{ \frac{9\pi}{2} \cdot (-2) \cdot 2 }{ \pi } = \frac{ \frac{9\pi}{2} \cdot (-4) }{ \pi } = \frac{ -18\pi }{ \pi } = -18. ]

So the answer is -18.

Let me verify with Python to avoid mistakes.

[PYTHON] import sympy as sp

Part (1): area

theta = sp.symbols('theta') r = 2 + sp.cos(theta) area_expr = sp.Rational(1,2) * sp.integrate(r**2, (theta, 0, 2*sp.pi)) print("Area a =", area_expr) print("a =", area_expr.simplify())

Part (2): slope at given point

theta_val = sp.pi/4 r_val = (4 + sp.sqrt(2))/2 dr_dtheta = -sp.sin(theta_val) # derivative of r w.r.t theta

dy/dtheta and dx/dtheta

dy_dtheta = dr_dtheta * sp.sin(theta_val) + r_val * sp.cos(theta_val) dx_dtheta = dr_dtheta * sp.cos(theta_val) - r_val * sp.sin(theta_val)

slope = sp.simplify(dy_dtheta / dx_dtheta) print("Slope =", slope)

Express as b + sqrt(c)

slope_simplified = sp.simplify(slope) print("Simplified slope =", slope_simplified)

Identify b and c: slope = b + sqrt(c)

slope = -2 + sqrt(2)

b = -2 c = 2 print("b =", b, "c =", c)

Compute abc/pi

a = 9*sp.pi/2 abc_over_pi = (a * b * c) / sp.pi print("abc/pi =", abc_over_pi) print("abc/pi as integer =", abc_over_pi.simplify()) [/PYTHON]

[THOUGHT] The Python output confirms: area ( a = \frac{9\pi}{2} ), slope = ( -2 + \sqrt{2} ) so ( b = -2 ), ( c = 2 ), and ( \frac{abc}{\pi} = -18 ). The final answer is (-18).

\boxed{-18}