The area enclosed by the curves xy + 4y = 16 and x + y = 6 is equal to :

Question

The area of the region enclosed between the circles $x^2 + y^2 = 4$ and $x^2 + (y - 2)^2 = 4$ is:

Answer 1

The provided equations are:

$ xy + 4y = 16 $ and $ x + y = 6 $

From the second equation, we isolate $ y $:

$ y = 6 - x $

Substitute this expression for $ y $ into the first equation:

$ x(6 - x) + 4(6 - x) = 16 $

Simplify the resulting equation:

$ 6x - x^2 + 24 - 4x = 16 $

$ 2x - x^2 = -8 $

$ x^2 - 2x - 8 = 0 $

Factor the quadratic equation to solve for $ x $:

$ (x - 4)(x + 2) = 0 $

The solutions for $ x $ are $ x = 4 $ and $ x = -2 $.

These values represent the limits of integration. The area between the curves is calculated by the following integral:

$ \text{Area} = \int_{-2}^{4} \left( \frac{16}{x + 4} - (6 - x) \right) \, dx $

Evaluating the integral yields the area:

$ \text{Area} = 30 - 32 \log 2 $

The Correct Option is C