If `(a,1/a), (b.1/b) , ( c,1/c) , (d , 1/d)` are four distinct points on a circle of radius 4 units then, `abcd` is equal to:

Options

A. 4
B. `1//4`
C. 1
D. 16

Answer

To solve the problem, we start with the given points \((a, \frac{1}{a}), (b, \frac{1}{b}), (c, \frac{1}{c}), (d, \frac{1}{d})\) which lie on a circle of radius 4 units. We need to find the product \(abcd\).

### Step-by-Step Solution:

1. **Equation of the Circle**:
The general equation of a circle with center at the origin and radius \(r\) is given by:
\[ x^2 + y^2 = r^2 \]
For a circle with radius 4, the equation becomes:
\[ x^2 + y^2 = 16 \]

2. **Substituting the Points**:
The points given are of the form \((x, y) = (x, \frac{1}{x})\). We substitute \(y = \frac{1}{x}\) into the circle's equation:
\[ x^2 + \left(\frac{1}{x}\right)^2 = 16 \]

3. **Simplifying the Equation**:
This simplifies to:
\[ x^2 + \frac{1}{x^2} = 16 \]
To eliminate the fraction, multiply through by \(x^2\):
\[ x^4 + 1 = 16x^2 \]

4. **Rearranging the Equation**:
Rearranging gives us a bi-quadratic equation:
\[ x^4 - 16x^2 + 1 = 0 \]

5. **Letting \(z = x^2\)**:
We can let \(z = x^2\), transforming our equation into:
\[ z^2 - 16z + 1 = 0 \]

6. **Using the Quadratic Formula**:
We can solve for \(z\) using the quadratic formula:
\[ z = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{16 \pm \sqrt{(-16)^2 - 4 \cdot 1 \cdot 1}}{2 \cdot 1} \]
\[ z = \frac{16 \pm \sqrt{256 - 4}}{2} = \frac{16 \pm \sqrt{252}}{2} = \frac{16 \pm 2\sqrt{63}}{2} = 8 \pm \sqrt{63} \]

7. **Finding \(x^2\)**:
The roots \(z_1\) and \(z_2\) are:
\[ z_1 = 8 + \sqrt{63}, \quad z_2 = 8 - \sqrt{63} \]
Hence, the corresponding \(x\) values are:
\[ x_1 = \sqrt{8 + \sqrt{63}}, \quad x_2 = -\sqrt{8 + \sqrt{63}}, \quad x_3 = \sqrt{8 - \sqrt{63}}, \quad x_4 = -\sqrt{8 - \sqrt{63}} \]

8. **Calculating the Product \(abcd\)**:
The product of the roots \(abcd\) can be calculated using the property of the roots of the bi-quadratic equation:
\[ abcd = z_1 z_2 = (8 + \sqrt{63})(8 - \sqrt{63}) = 8^2 - (\sqrt{63})^2 = 64 - 63 = 1 \]

### Final Answer:
Thus, the value of \(abcd\) is:
\[ \boxed{1} \]