Find the smaller area enclosed by the circle `x ^(2) + y ^(2) =4 ` and the line `x + y =2.`

Answer

To find the smaller area enclosed by the circle \(x^2 + y^2 = 4\) and the line \(x + y = 2\), we will follow these steps:

### Step 1: Identify the equations
The equation of the circle is given by:
\[ x^2 + y^2 = 4 \]
This represents a circle centered at the origin (0,0) with a radius of 2.

The equation of the line is:
\[ x + y = 2 \]
We can rewrite this in slope-intercept form as:
\[ y = 2 - x \]

### Step 2: Find the points of intersection
To find the points where the line intersects the circle, we substitute \(y = 2 - x\) into the circle's equation:
\[ x^2 + (2 - x)^2 = 4 \]
Expanding this:
\[ x^2 + (4 - 4x + x^2) = 4 \]
\[ 2x^2 - 4x + 4 - 4 = 0 \]
\[ 2x^2 - 4x = 0 \]
Factoring out \(2x\):
\[ 2x(x - 2) = 0 \]
This gives us \(x = 0\) and \(x = 2\).

Now, substituting these \(x\) values back into the line equation to find corresponding \(y\) values:
- For \(x = 0\): \(y = 2\) → Point (0, 2)
- For \(x = 2\): \(y = 0\) → Point (2, 0)

### Step 3: Set up the integral
The area enclosed between the curves can be found by integrating the difference between the upper curve (the circle) and the lower curve (the line) from \(x = 0\) to \(x = 2\).

The equation of the circle can be expressed as:
\[ y = \sqrt{4 - x^2} \]
Thus, the area \(A\) can be expressed as:
\[ A = \int_{0}^{2} \left(\sqrt{4 - x^2} - (2 - x)\right) \, dx \]

### Step 4: Evaluate the integral
Now we evaluate the integral:
\[ A = \int_{0}^{2} \left(\sqrt{4 - x^2} - 2 + x\right) \, dx \]

We can split this into two integrals:
\[ A = \int_{0}^{2} \sqrt{4 - x^2} \, dx - \int_{0}^{2} (2 - x) \, dx \]

1. **Evaluate \(\int_{0}^{2} \sqrt{4 - x^2} \, dx\)**:
This integral represents the area of a quarter circle with radius 2:
\[ \int_{0}^{2} \sqrt{4 - x^2} \, dx = \frac{1}{4} \pi (2^2) = \pi \]

2. **Evaluate \(\int_{0}^{2} (2 - x) \, dx\)**:
\[ \int_{0}^{2} (2 - x) \, dx = \left[2x - \frac{x^2}{2}\right]_{0}^{2} = \left[4 - 2\right] - [0] = 2 \]

### Step 5: Combine the results
Now, substituting back into the area formula:
\[ A = \pi - 2 \]

### Conclusion
The smaller area enclosed by the circle and the line is:
\[ \boxed{\pi - 2} \]