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

Answer

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

### Step-by-Step Solution:

1. **Identify the Circle**: The given equation \( x^2 + y^2 = 1 \) represents a circle with center at the origin (0, 0) and radius 1.

2. **Equation of the Tangent**: The general equation of a tangent to the circle \( x^2 + y^2 = r^2 \) at a point \( (x_1, y_1) \) on the circle is given by:
\[ xx_1 + yy_1 = r^2 \]
For our circle, \( r^2 = 1 \). Thus, the equation becomes:
\[ xx_1 + yy_1 = 1 \]

3. **Intercepts on the Axes**: The tangent line will intersect the x-axis and y-axis at points P and Q. To find these intercepts, we can set \( y = 0 \) for the x-intercept and \( x = 0 \) for the y-intercept.

- **X-intercept (P)**: Set \( y = 0 \):
\[ xx_1 = 1 \implies x = \frac{1}{x_1} \quad \text{(Let this point be } P\left(\frac{1}{x_1}, 0\right)\text{)} \]

- **Y-intercept (Q)**: Set \( x = 0 \):
\[ yy_1 = 1 \implies y = \frac{1}{y_1} \quad \text{(Let this point be } Q\left(0, \frac{1}{y_1}\right)\text{)} \]

4. **Midpoint of PQ**: The midpoint \( M \) of the points \( P \) and \( Q \) is given by:
\[ M\left(\frac{\frac{1}{x_1} + 0}{2}, \frac{0 + \frac{1}{y_1}}{2}\right) = M\left(\frac{1}{2x_1}, \frac{1}{2y_1}\right) \]

5. **Relating \( x_1 \) and \( y_1 \)**: Since \( (x_1, y_1) \) lies on the circle, we have:
\[ x_1^2 + y_1^2 = 1 \]

6. **Expressing \( x_1 \) and \( y_1 \)** in terms of \( h \) and \( k \) (the coordinates of the midpoint):
- From \( M\left(h, k\right) = \left(\frac{1}{2x_1}, \frac{1}{2y_1}\right) \), we can express:
\[ x_1 = \frac{1}{2h}, \quad y_1 = \frac{1}{2k} \]

7. **Substituting into the Circle Equation**:
\[ \left(\frac{1}{2h}\right)^2 + \left(\frac{1}{2k}\right)^2 = 1 \]
Simplifying this gives:
\[ \frac{1}{4h^2} + \frac{1}{4k^2} = 1 \implies \frac{1}{h^2} + \frac{1}{k^2} = 4 \]

8. **Finding the Locus**: Rearranging gives:
\[ \frac{k^2 + h^2}{h^2k^2} = 4 \implies k^2 + h^2 = 4h^2k^2 \]

This can be rewritten as:
\[ h^2 + k^2 - 4h^2k^2 = 0 \]

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