If a tangent to the circle `x^(2)+y^(2)=1` intersects the coordinate axes at distinct points P and Q, then the locus of the mid-point of PQ is :

Options

A. `x^(2)+y^(2)-2xy=0`
B. `x^(2)+y^(2)-16x^(2)y^(2)=0`
C. `x^(2)+y^(2)-2x^(2)y^(2)=0`
D. `x^(2)+y^(2)-4x^(2)y^(2)=0`

Explanation

To find the locus of the midpoint of the line segment PQ, where P and Q are the points where a tangent to the circle \(x^2 + y^2 = 1\) intersects the coordinate axes, we can follow these steps:

### Step 1: Understand the Circle and Tangent
The given circle is centered at the origin with a radius of 1. The equation of the circle is:
\[ x^2 + y^2 = 1 \]
A tangent to this circle can be expressed in the form:
\[ xx_1 + yy_1 = 1 \]
where \((x_1, y_1)\) is a point on the circle.

### Step 2: Parametrize the Point on the Circle
We can use parametric coordinates for the point on the circle:
\[ x_1 = \cos \theta, \quad y_1 = \sin \theta \]
Thus, the equation of the tangent becomes:
\[ x \cos \theta + y \sin \theta = 1 \]

### Step 3: Find Intersection Points with Axes
To find the intersection points P and Q with the axes, we set \(y = 0\) to find point Q and \(x = 0\) to find point P.

1. **Finding Point P (x = 0)**:
\[ 0 \cdot \cos \theta + y \sin \theta = 1 \implies y = \frac{1}{\sin \theta} \implies P = \left(0, \csc \theta\right) \]

2. **Finding Point Q (y = 0)**:
\[ x \cos \theta + 0 \cdot \sin \theta = 1 \implies x = \frac{1}{\cos \theta} \implies Q = \left(\sec \theta, 0\right) \]

### Step 4: Midpoint of PQ
The midpoint R of the segment PQ can be calculated as:
\[ R = \left(\frac{0 + \sec \theta}{2}, \frac{\csc \theta + 0}{2}\right) = \left(\frac{\sec \theta}{2}, \frac{\csc \theta}{2}\right) \]

### Step 5: Express Midpoint Coordinates in Terms of h and k
Let:
\[ h = \frac{\sec \theta}{2}, \quad k = \frac{\csc \theta}{2} \]
From these, we can express:
\[ \sec \theta = 2h \quad \text{and} \quad \csc \theta = 2k \]

### Step 6: Use Trigonometric Identities
Using the identity \(\sec^2 \theta - \tan^2 \theta = 1\) and \(\csc^2 \theta - \cot^2 \theta = 1\), we can relate \(h\) and \(k\):
\[ \sec^2 \theta + \csc^2 \theta = 1 + \tan^2 \theta + 1 + \cot^2 \theta \]
Substituting \(2h\) and \(2k\):
\[ (2h)^2 + (2k)^2 = 4h^2 + 4k^2 = 4 \]
Thus, we have:
\[ h^2 + k^2 = 1 \]

### Step 7: Final Locus Equation
The locus of the midpoint R, which is \((h, k)\), can be expressed as:
\[ h^2 + k^2 = 1 \]
Replacing \(h\) and \(k\) with \(x\) and \(y\):
\[ x^2 + y^2 = 1 \]

However, we need to account for the factor of 4 introduced earlier. The final locus equation becomes:
\[ x^2 + y^2 - 4 = 0 \]

### Conclusion
The locus of the midpoint of PQ is given by:
\[ x^2 + y^2 - 4 = 0 \]