A point P lying inside the curve `y = sqrt(2ax-x^2)` is moving such that its shortest distance from the curve at any position is greater than its distance from X-axis. The point P enclose a region whose area is equal to

Options

A. (a)`(pi a^2)/2`
B. (b)`(a^2)/(3)`
C. (c)`(2a^2)/(3)`
D. (d)`((3pi-4)/(6))a^2`

Answer

To solve the problem, we need to find the area enclosed by the point P that lies inside the curve given by the equation \( y = \sqrt{2ax - x^2} \) under the condition that the shortest distance from the curve is greater than its distance from the x-axis.

### Step-by-Step Solution:

1. **Understand the Curve**:
The equation \( y = \sqrt{2ax - x^2} \) can be rewritten by squaring both sides:
\[ y^2 = 2ax - x^2 \]
Rearranging gives:
\[ x^2 + y^2 - 2ax = 0 \]

2. **Complete the Square**:
We can complete the square for the \( x \) terms:
\[ (x - a)^2 + y^2 = a^2 \]
This represents a circle with center at \( (a, 0) \) and radius \( a \).

3. **Distance Conditions**:
The shortest distance from the point \( P(h, k) \) to the curve must be greater than the distance from \( P \) to the x-axis. The distance from \( P \) to the x-axis is simply \( k \), and the distance from \( P \) to the curve can be expressed using the distance formula.

4. **Setting Up the Inequality**:
We need to express the condition mathematically:
\[ d(P, \text{curve}) > k \]
The distance from point \( P(h, k) \) to the curve can be derived from the geometry of the circle.

5. **Using Geometry**:
The distance from point \( P(h, k) \) to the center of the circle \( (a, 0) \) is:
\[ d = \sqrt{(h - a)^2 + k^2} \]
For the point \( P \) to be inside the circle, we need:
\[ \sqrt{(h - a)^2 + k^2} < a \]
The condition becomes:
\[ \sqrt{(h - a)^2 + k^2} > k \]

6. **Squaring Both Sides**:
Squaring both sides of the inequality gives:
\[ (h - a)^2 + k^2 > k^2 \]
Which simplifies to:
\[ (h - a)^2 > 0 \]
This means \( h \neq a \).

7. **Finding the Area**:
The area enclosed by point \( P \) can be calculated as follows:
The area of the circle is given by:
\[ \text{Area} = \pi r^2 = \pi a^2 \]
However, we need the area where \( P \) satisfies the conditions above. The area we are interested in is half of the circle, as \( P \) must be above the x-axis.

8. **Final Calculation**:
The area of the region enclosed by \( P \) is:
\[ \text{Area} = \frac{1}{2} \pi a^2 \]

### Conclusion:
The area enclosed by point \( P \) is \( \frac{a^2}{2} \).