To solve the problem, we need to find the area enclosed by the locus of the center of a variable circle that touches the line \(y = 0\) and passes through the point \((0, 1)\), and the line \(x + y = 2\).
### Step-by-Step Solution:
1. **Understanding the Circle**:
- Let the center of the circle be \((h, k)\).
- Since the circle touches the line \(y = 0\) (the x-axis), the radius \(r\) of the circle is equal to the y-coordinate of the center, which means \(r = k\).
2. **Using the Point on the Circle**:
- The circle passes through the point \((0, 1)\). Therefore, the equation of the circle can be expressed as:
\[
(0 - h)^2 + (1 - k)^2 = k^2
\]
- Simplifying this gives:
\[
h^2 + (1 - k)^2 = k^2
\]
- Expanding the left side:
\[
h^2 + 1 - 2k + k^2 = k^2
\]
- Canceling \(k^2\) from both sides:
\[
h^2 + 1 - 2k = 0
\]
- Rearranging gives the locus of the center:
\[
h^2 = 2k - 1
\]
3. **Substituting for Locus**:
- Replacing \(h\) with \(x\) and \(k\) with \(y\), we have:
\[
x^2 = 2y - 1
\]
- This can be rewritten as:
\[
y = \frac{x^2 + 1}{2}
\]
- This is the equation of the parabola \(P\).
4. **Finding the Area Enclosed by the Curve and the Line**:
- The line \(x + y = 2\) can be rewritten as:
\[
y = 2 - x
\]
- We need to find the points of intersection between the parabola and the line:
\[
\frac{x^2 + 1}{2} = 2 - x
\]
- Multiplying through by 2 to eliminate the fraction:
\[
x^2 + 1 = 4 - 2x
\]
- Rearranging gives:
\[
x^2 + 2x - 3 = 0
\]
- Factoring:
\[
(x + 3)(x - 1) = 0
\]
- Thus, the solutions are \(x = -3\) and \(x = 1\).
5. **Setting Up the Integral for Area**:
- The area \(A\) between the curves from \(x = -3\) to \(x = 1\) is given by:
\[
A = \int_{-3}^{1} \left[(2 - x) - \left(\frac{x^2 + 1}{2}\right)\right] \, dx
\]
- Simplifying the integrand:
\[
A = \int_{-3}^{1} \left(2 - x - \frac{x^2}{2} - \frac{1}{2}\right) \, dx = \int_{-3}^{1} \left(\frac{3}{2} - x - \frac{x^2}{2}\right) \, dx
\]
6. **Calculating the Integral**:
- Break it down:
\[
A = \int_{-3}^{1} \frac{3}{2} \, dx - \int_{-3}^{1} x \, dx - \int_{-3}^{1} \frac{x^2}{2} \, dx
\]
- Evaluating each integral:
- \(\int_{-3}^{1} \frac{3}{2} \, dx = \frac{3}{2} \cdot (1 - (-3)) = \frac{3}{2} \cdot 4 = 6\)
- \(\int_{-3}^{1} x \, dx = \left[\frac{x^2}{2}\right]_{-3}^{1} = \frac{1^2}{2} - \frac{(-3)^2}{2} = \frac{1}{2} - \frac{9}{2} = -4\)
- \(\int_{-3}^{1} \frac{x^2}{2} \, dx = \frac{1}{2} \left[\frac{x^3}{3}\right]_{-3}^{1} = \frac{1}{2} \left(\frac{1}{3} - \frac{-27}{3}\right) = \frac{1}{2} \cdot \frac{28}{3} = \frac{14}{3}\)
7. **Combining the Results**:
- Thus, the area \(A\) becomes:
\[
A = 6 - (-4) - \frac{14}{3} = 6 + 4 - \frac{14}{3} = 10 - \frac{14}{3} = \frac{30}{3} - \frac{14}{3} = \frac{16}{3}
\]
### Final Answer:
The area enclosed by the curve \(P\) and the line \(x + y = 2\) is \(\frac{16}{3}\) square units.
