Locus of middle point of intercept of any tangent with respect to the circle `x^(2) + y^(2) =4` between the axis is

Options

A. `x^(2) + y^(2) - x^(2)y^(2) =0`
B. `x^(2) + y^(2) + x^(2) y^(2) =0`
C. `x^(2) + y^(2) -2x^(2) y^(2) =0`
D. `x^(2) + y^(2) - 3x^(2)y^(2) =0`

Answer

To find the locus of the midpoint of the intercept of any tangent with respect to the circle \(x^2 + y^2 = 4\) between the axes, we can follow these steps:

### Step 1: Understand the Circle
The given equation of the circle is \(x^2 + y^2 = 4\). This represents a circle centered at the origin \((0, 0)\) with a radius of \(2\).

### Step 2: Equation of the Tangent
The general equation of a tangent to the circle \(x^2 + y^2 = r^2\) (where \(r = 2\)) can be expressed as:
\[ y = mx + \sqrt{r^2(1 + m^2)} \]
Substituting \(r = 2\):
\[ y = mx + 2\sqrt{1 + m^2} \]

### Step 3: Find Intercepts
To find the intercepts of the tangent with the axes:
- **Y-intercept (Point B)**: Set \(x = 0\):
\[ y = 2\sqrt{1 + m^2} \]
Thus, the coordinates of point B are \((0, 2\sqrt{1 + m^2})\).

- **X-intercept (Point A)**: Set \(y = 0\):
\[ 0 = mx + 2\sqrt{1 + m^2} \implies mx = -2\sqrt{1 + m^2} \implies x = -\frac{2\sqrt{1 + m^2}}{m} \]
Thus, the coordinates of point A are \(\left(-\frac{2\sqrt{1 + m^2}}{m}, 0\right)\).

### Step 4: Midpoint of AB
Let \(M\) be the midpoint of \(AB\). The coordinates of \(M\) are given by:
\[ M\left(h, k\right) = \left(\frac{x_A + x_B}{2}, \frac{y_A + y_B}{2}\right) \]
Substituting the coordinates of points A and B:
\[ h = \frac{-\frac{2\sqrt{1 + m^2}}{m} + 0}{2} = -\frac{\sqrt{1 + m^2}}{m} \]
\[ k = \frac{0 + 2\sqrt{1 + m^2}}{2} = \sqrt{1 + m^2} \]

### Step 5: Eliminate \(m\)
From the expressions for \(h\) and \(k\):
1. \(h = -\frac{\sqrt{1 + m^2}}{m}\)
2. \(k = \sqrt{1 + m^2}\)

Squaring both sides:
- From \(k\):
\[ k^2 = 1 + m^2 \implies m^2 = k^2 - 1 \]

- Substitute \(m^2\) in the equation for \(h\):
\[ h^2 = \frac{1 + m^2}{m^2} = \frac{1 + (k^2 - 1)}{k^2 - 1} = \frac{k^2}{k^2 - 1} \]

Cross-multiplying gives:
\[ h^2(k^2 - 1) = k^2 \implies h^2k^2 - h^2 = k^2 \implies h^2 + k^2 = h^2k^2 \]

### Step 6: Final Locus Equation
Rearranging gives the final locus equation:
\[ x^2 + y^2 = x^2y^2 \]

### Conclusion
The locus of the midpoint of the intercept of any tangent with respect to the circle \(x^2 + y^2 = 4\) is given by the equation:
\[ x^2 + y^2 = x^2y^2 \]