A line `L` varies such that length of perpendicular on it from origin `O` is always 4 units. If `L` cuts x-axis and y-axis at `A` and `B` respectively then minimum value of `(OA)^(2)+(OB)^(2)` is

Options

A. `16`
B. `32`
C. `64`
D. `128`

Explanation

To solve the problem, we need to find the minimum value of \( OA^2 + OB^2 \) where \( O \) is the origin, \( A \) is the point where the line intersects the x-axis, and \( B \) is the point where the line intersects the y-axis. The line \( L \) is such that the length of the perpendicular from the origin to the line is always 4 units.

### Step-by-Step Solution:

1. **Equation of the Line**:
The equation of a line in terms of the perpendicular distance from the origin can be expressed as:
\[ x \cos \alpha + y \sin \alpha = d \]
where \( d \) is the distance from the origin to the line. Here, \( d = 4 \), so the equation becomes:
\[ x \cos \alpha + y \sin \alpha = 4 \]

2. **Finding Points A and B**:
- To find the x-intercept \( A \) (where \( y = 0 \)):
\[ x \cos \alpha = 4 \implies x = \frac{4}{\cos \alpha} \implies A\left(\frac{4}{\cos \alpha}, 0\right) \]
- To find the y-intercept \( B \) (where \( x = 0 \)):
\[ y \sin \alpha = 4 \implies y = \frac{4}{\sin \alpha} \implies B\left(0, \frac{4}{\sin \alpha}\right) \]

3. **Calculating Distances OA and OB**:
- The distance \( OA \) is:
\[ OA = \frac{4}{\cos \alpha} \]
- The distance \( OB \) is:
\[ OB = \frac{4}{\sin \alpha} \]

4. **Finding \( OA^2 + OB^2 \)**:
\[ OA^2 = \left(\frac{4}{\cos \alpha}\right)^2 = \frac{16}{\cos^2 \alpha} \]
\[ OB^2 = \left(\frac{4}{\sin \alpha}\right)^2 = \frac{16}{\sin^2 \alpha} \]
Therefore,
\[ OA^2 + OB^2 = \frac{16}{\cos^2 \alpha} + \frac{16}{\sin^2 \alpha} = 16\left(\frac{1}{\cos^2 \alpha} + \frac{1}{\sin^2 \alpha}\right) \]

5. **Using Trigonometric Identities**:
We know that:
\[ \frac{1}{\cos^2 \alpha} = \sec^2 \alpha \quad \text{and} \quad \frac{1}{\sin^2 \alpha} = \csc^2 \alpha \]
Thus,
\[ OA^2 + OB^2 = 16(\sec^2 \alpha + \csc^2 \alpha) \]

6. **Finding Minimum Value**:
Using the identity \( \sec^2 \alpha + \csc^2 \alpha = 1 + \tan^2 \alpha + 1 + \cot^2 \alpha \):
\[ \sec^2 \alpha + \csc^2 \alpha = 2 + \tan^2 \alpha + \cot^2 \alpha \]
The minimum value of \( \tan^2 \alpha + \cot^2 \alpha \) occurs at \( \tan \alpha = 1 \) (i.e., \( \alpha = 45^\circ \)), giving:
\[ \tan^2 \alpha + \cot^2 \alpha = 2 \]
Therefore,
\[ \sec^2 \alpha + \csc^2 \alpha = 4 \]
Hence,
\[ OA^2 + OB^2 = 16 \times 4 = 64 \]

### Conclusion:
The minimum value of \( OA^2 + OB^2 \) is \( 64 \).