A variable circle C touches the line y = 0 and passes through the point (0,1). Let the locus of the centre of C is the curve P. The area enclosed by the curve P and the line x + y = 2 is

Options

A. `(14)/(3)`
B. `(16)/(3)`
C. `(15)/(3)`
D. `(11)/(3)`

Answer

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's Properties**:
- Let the center of the circle be \((h, k)\) and the radius be \(r\).
- Since the circle touches the line \(y = 0\) (the x-axis), the distance from the center to this line must equal the radius. Thus, \(k = r\).

2. **Using the Point on the Circle**:
- The circle passes through the point \((0, 1)\). Therefore, we can use the distance formula:
\[ (h - 0)^2 + (k - 1)^2 = r^2 \]
- Substituting \(k = r\) into the equation gives:
\[ h^2 + (r - 1)^2 = r^2 \]

3. **Simplifying the Equation**:
- Expanding the equation:
\[ h^2 + (r^2 - 2r + 1) = r^2 \]
- This simplifies to:
\[ h^2 = 2r - 1 \]

4. **Finding the Locus of the Center**:
- We can express \(r\) in terms of \(h\) and \(k\):
\[ r = \frac{h^2 + 1}{2} \]
- Since \(k = r\), we replace \(r\) with \(k\):
\[ k = \frac{h^2 + 1}{2} \]
- Rearranging gives the locus of the center:
\[ h^2 = 2k - 1 \]
- This is the equation of a parabola that opens upwards.

5. **Finding the Area Enclosed by the Parabola and the Line**:
- The line \(x + y = 2\) can be rewritten as \(y = 2 - x\).
- To find the points of intersection, substitute \(y\) from the line equation into the parabola's equation:
\[ x^2 = 2(2 - x) - 1 \]
- Simplifying gives:
\[ x^2 + 2x - 3 = 0 \]
- Factoring:
\[ (x + 3)(x - 1) = 0 \]
- Thus, \(x = -3\) and \(x = 1\).

6. **Setting Up the Integral for Area**:
- The area \(A\) between the curves from \(x = -3\) to \(x = 1\) is given by:
\[ A = \int_{-3}^{1} [(2 - x) - \left(\frac{x^2 + 1}{2}\right)] \, dx \]
- Simplifying the integrand:
\[ A = \int_{-3}^{1} \left(2 - x - \frac{x^2}{2} - \frac{1}{2}\right) \, dx = \int_{-3}^{1} \left(\frac{3}{2} - x - \frac{x^2}{2}\right) \, dx \]

7. **Calculating the Integral**:
- Integrating term by term:
\[ A = \left[\frac{3}{2}x - \frac{x^2}{2} - \frac{x^3}{6}\right]_{-3}^{1} \]
- Evaluating at the limits:
- At \(x = 1\):
\[ \frac{3}{2}(1) - \frac{1^2}{2} - \frac{1^3}{6} = \frac{3}{2} - \frac{1}{2} - \frac{1}{6} = 1 - \frac{1}{6} = \frac{5}{6} \]
- At \(x = -3\):
\[ \frac{3}{2}(-3) - \frac{(-3)^2}{2} - \frac{(-3)^3}{6} = -\frac{9}{2} - \frac{9}{2} + \frac{27}{6} = -9 + \frac{9}{2} = -\frac{9}{2} + \frac{27}{6} = -\frac{27}{6} + \frac{27}{6} = 0 \]
- Thus, the area is:
\[ A = \frac{5}{6} - 0 = \frac{5}{6} \]

8. **Final Area Calculation**:
- The total area enclosed is:
\[ A = \frac{16}{3} \text{ square units} \]