A chord of the circle `x^(2)+y^(2)=a^(2)` cuts it at two points A and B such that `angle AOB = pi //2`, where O is the centre of the circle. If there is a moving point P on this circle, then the locus of the orthocentre of `DeltaPAB` will be a

Options

A. parabola
B. circle
C. straight line
D. none of these

Explanation

To solve the problem, we need to find the locus of the orthocenter of triangle \( PAB \) where \( P \) is a moving point on the circle defined by the equation \( x^2 + y^2 = a^2 \), and \( A \) and \( B \) are points on the circle such that \( \angle AOB = \frac{\pi}{2} \).

### Step-by-Step Solution:

1. **Identify the Circle and Points**:
The equation of the circle is given by:
\[ x^2 + y^2 = a^2 \]
The center of the circle \( O \) is at the origin \( (0, 0) \).

2. **Determine Points A and B**:
Since \( \angle AOB = \frac{\pi}{2} \), points \( A \) and \( B \) are at right angles to each other. We can choose:
\[ A = (a, 0) \quad \text{and} \quad B = (0, a) \]

3. **Moving Point P**:
Let \( P \) be a point on the circle with coordinates \( (h, k) \) such that:
\[ h^2 + k^2 = a^2 \]

4. **Find the Orthocenter of Triangle PAB**:
The orthocenter \( H \) of triangle \( PAB \) can be found using the properties of the triangle. The circumcenter \( C \) of triangle \( PAB \) is at the origin \( O(0, 0) \).

5. **Centroid of Triangle PAB**:
The coordinates of the centroid \( G \) of triangle \( PAB \) are given by:
\[ G = \left( \frac{h + a + 0}{3}, \frac{k + 0 + a}{3} \right) = \left( \frac{h + a}{3}, \frac{k + a}{3} \right) \]

6. **Relationship Between Orthocenter, Centroid, and Circumcenter**:
The centroid \( G \) divides the line segment joining the orthocenter \( H \) and the circumcenter \( C \) in the ratio \( 2:1 \):
\[ G = \frac{2H + C}{3} \]
Since \( C \) is at the origin \( (0, 0) \), we have:
\[ G = \frac{2H}{3} \]

7. **Setting Up the Equations**:
From the coordinates of \( G \):
\[ \frac{h + a}{3} = \frac{2\alpha}{3} \quad \text{and} \quad \frac{k + a}{3} = \frac{2\beta}{3} \]
This gives us:
\[ h + a = 2\alpha \quad \text{and} \quad k + a = 2\beta \]
Rearranging gives:
\[ h = 2\alpha - a \quad \text{and} \quad k = 2\beta - a \]

8. **Substituting into the Circle Equation**:
Substitute \( h \) and \( k \) into the equation of the circle:
\[ (2\alpha - a)^2 + (2\beta - a)^2 = a^2 \]
Expanding this:
\[ 4\alpha^2 - 4a\alpha + a^2 + 4\beta^2 - 4a\beta + a^2 = a^2 \]
Simplifying gives:
\[ 4\alpha^2 + 4\beta^2 - 4a(\alpha + \beta) + a^2 = 0 \]
Rearranging leads to the equation:
\[ \alpha^2 + \beta^2 - a(\alpha + \beta) + \frac{a^2}{4} = 0 \]

9. **Identifying the Locus**:
This equation represents a circle with center at \( (a/2, a/2) \) and radius \( a \).

### Conclusion:
The locus of the orthocenter of triangle \( PAB \) as point \( P \) moves around the circle \( x^2 + y^2 = a^2 \) is a circle centered at \( (a, a) \) with radius \( a \).