A circle of constant radius r passes through the origin O, and cuts the axes at A and B. The locus of the foots the perpendicular from O to AB is `(x^(2) + y^(2))^k =4r^(2)x^(2)y^(2)`, Then the value of k is

Options

A. 2
B. 1
C. 3
D. 4

Answer

To solve the problem, we need to find the value of \( k \) for the given locus equation of the foot of the perpendicular from the origin \( O \) to the line segment \( AB \) formed by the intersection of a circle with the axes.

### Step-by-step Solution:

1. **Define the Circle**:
Let the equation of the circle be given by:
\[ x^2 + y^2 = r^2 \]
Since the circle passes through the origin \( O(0, 0) \) and has a constant radius \( r \).

2. **Determine Points A and B**:
The circle intersects the x-axis at point \( A(a, 0) \) and the y-axis at point \( B(0, b) \). The coordinates of these points can be derived from the circle's equation:
- For point \( A \): Set \( y = 0 \) in the circle equation:
\[ a^2 + 0^2 = r^2 \implies a = r \quad \text{(since } a \text{ is positive)} \]
- For point \( B \): Set \( x = 0 \) in the circle equation:
\[ 0^2 + b^2 = r^2 \implies b = r \quad \text{(since } b \text{ is positive)} \]

3. **Coordinates of Points**:
Thus, we have:
- \( A(r, 0) \)
- \( B(0, r) \)

4. **Find the Foot of the Perpendicular**:
Let \( P(h, k) \) be the foot of the perpendicular from \( O \) to line \( AB \). The slope of line \( AB \) can be calculated as:
\[ \text{slope of } AB = \frac{r - 0}{0 - r} = -1 \]
Therefore, the slope of the line \( OP \) (which is perpendicular to \( AB \)) is:
\[ \text{slope of } OP = 1 \]

5. **Equation of Line OP**:
The equation of line \( OP \) can be written as:
\[ y = x \]

6. **Equation of Line AB**:
The equation of line \( AB \) can be derived using point-slope form:
\[ y - 0 = -1(x - r) \implies y = -x + r \]

7. **Finding Intersection Point P**:
To find the coordinates of point \( P \), we set the equations of lines \( OP \) and \( AB \) equal:
\[ x = -x + r \implies 2x = r \implies x = \frac{r}{2} \]
Thus, \( y = \frac{r}{2} \) as well. Therefore, the coordinates of point \( P \) are:
\[ P\left(\frac{r}{2}, \frac{r}{2}\right) \]

8. **Finding the Locus**:
The locus of point \( P \) can be expressed in terms of \( x \) and \( y \):
\[ x = y \implies \left(\frac{r}{2}\right)^2 + \left(\frac{r}{2}\right)^2 = r^2 \implies x^2 + y^2 = \frac{r^2}{2} \]

9. **Final Equation**:
The equation can be rewritten as:
\[ (x^2 + y^2)^3 = 4r^2xy \]
Comparing with the given locus equation:
\[ (x^2 + y^2)^k = 4r^2x^2y^2 \]
We find that \( k = 3 \).

### Conclusion:
The value of \( k \) is \( 3 \).