If `AB` is the intercept of the tangent to the circle `x^2 +y^2=r^2` between the coordinate axes, the locus of the vertex `P` of the rectangle `OAPB` is

Options

A. `x^(2)+y^(2)=r^(2)`
B. `(1)/(x^(2))+(1)/(y^(2))=(1)/(r^(2))`
C. `(1)/(x^(2))+(1)/(y^(2))=r^(2)`
D. none of these

Explanation

To solve the problem, we need to find the locus of the vertex \( P \) of the rectangle \( OAPB \) formed by the intercepts \( A \) and \( B \) of the tangent to the circle \( x^2 + y^2 = r^2 \).

### Step-by-Step Solution:

1. **Equation of the Circle**:
The equation of the circle is given by:
\[ x^2 + y^2 = r^2 \]

2. **Equation of the Tangent**:
The equation of the tangent to the circle at the point \( (r \cos \theta, r \sin \theta) \) can be written as:
\[ r \cos \theta \cdot x + r \sin \theta \cdot y = r^2 \]
This can be simplified to:
\[ x \cos \theta + y \sin \theta = r \]

3. **Finding Intercepts**:
- **Intercept on the x-axis (Point A)**: Set \( y = 0 \):
\[ x \cos \theta = r \implies x = \frac{r}{\cos \theta} \implies A\left(\frac{r}{\cos \theta}, 0\right) \]
- **Intercept on the y-axis (Point B)**: Set \( x = 0 \):
\[ y \sin \theta = r \implies y = \frac{r}{\sin \theta} \implies B\left(0, \frac{r}{\sin \theta}\right) \]

4. **Coordinates of Points A and B**:
- \( A = \left(r \sec \theta, 0\right) \)
- \( B = \left(0, r \csc \theta\right) \)

5. **Coordinates of Point P**:
The coordinates of point \( P \) (the vertex of rectangle \( OAPB \)) are:
\[ P = (x, y) = \left(r \sec \theta, r \csc \theta\right) \]

6. **Relating Coordinates to r**:
From the coordinates of \( P \):
\[ h = r \sec \theta \quad \text{and} \quad k = r \csc \theta \]

7. **Using Trigonometric Identities**:
We know:
\[ \sec^2 \theta + \csc^2 \theta = 1 + \tan^2 \theta + 1 + \cot^2 \theta \]
However, we can also use:
\[ \frac{1}{h^2} + \frac{1}{k^2} = \frac{1}{r^2} \]

8. **Final Locus Equation**:
Replacing \( h \) and \( k \) with \( x \) and \( y \):
\[ \frac{1}{x^2} + \frac{1}{y^2} = \frac{1}{r^2} \]

### Conclusion:
The locus of the vertex \( P \) of the rectangle \( OAPB \) is given by:
\[ \frac{1}{x^2} + \frac{1}{y^2} = \frac{1}{r^2} \]