The locus of the middle points of portions of the tangents to the circle `x^(2)+y^(2)=a^(2)` terminated by the axes is

Options

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

Answer

To find the locus of the midpoints of the portions of the tangents to the circle \(x^2 + y^2 = a^2\) that are terminated by the axes, we will follow these steps:

### Step 1: Equation of the Tangent
The equation of the tangent to the circle \(x^2 + y^2 = a^2\) at a point \((x_1, y_1)\) on the circle is given by:
\[ x_1 x + y_1 y = a^2 \]

### Step 2: Parametrizing the Point on the Circle
We can parametrize the point \((x_1, y_1)\) on the circle using trigonometric functions:
\[ x_1 = a \cos \theta, \quad y_1 = a \sin \theta \]

### Step 3: Substitute into the Tangent Equation
Substituting \(x_1\) and \(y_1\) into the tangent equation gives:
\[ a \cos \theta \cdot x + a \sin \theta \cdot y = a^2 \]
Dividing through by \(a\) (assuming \(a \neq 0\)):
\[ \cos \theta \cdot x + \sin \theta \cdot y = a \]

### Step 4: Finding Intercepts
To find the intercepts on the axes:
- The x-intercept occurs when \(y = 0\):
\[ x = \frac{a}{\cos \theta} \quad \text{(let this be } L\text{)} \]
- The y-intercept occurs when \(x = 0\):
\[ y = \frac{a}{\sin \theta} \quad \text{(let this be } M\text{)} \]

### Step 5: Midpoint of the Segment
The midpoint \( (H, K) \) of the segment \(AB\) formed by the intercepts \(L\) and \(M\) is given by:
\[ H = \frac{L + 0}{2} = \frac{a}{2 \cos \theta}, \quad K = \frac{M + 0}{2} = \frac{a}{2 \sin \theta} \]

### Step 6: Expressing in Terms of \(H\) and \(K\)
From the midpoint expressions, we have:
\[ 2H = \frac{a}{\cos \theta} \quad \Rightarrow \quad \cos \theta = \frac{a}{2H} \]
\[ 2K = \frac{a}{\sin \theta} \quad \Rightarrow \quad \sin \theta = \frac{a}{2K} \]

### Step 7: Using the Pythagorean Identity
Using the identity \(\cos^2 \theta + \sin^2 \theta = 1\):
\[ \left(\frac{a}{2H}\right)^2 + \left(\frac{a}{2K}\right)^2 = 1 \]
This simplifies to:
\[ \frac{a^2}{4H^2} + \frac{a^2}{4K^2} = 1 \]

### Step 8: Rearranging the Equation
Multiplying through by \(4H^2K^2\) gives:
\[ a^2 K^2 + a^2 H^2 = 4H^2K^2 \]
Rearranging leads to:
\[ \frac{a^2}{4H^2} + \frac{a^2}{4K^2} = 1 \]
This can be rewritten as:
\[ \frac{a^2}{4H^2} + \frac{a^2}{4K^2} = 1 \]
or:
\[ \frac{a^2}{4x^2} + \frac{a^2}{4y^2} = 1 \]

### Step 9: Final Form of the Locus
Multiplying through by \(4\) gives:
\[ \frac{a^2}{x^2} + \frac{a^2}{y^2} = 4 \]
Thus, the locus of the midpoints is:
\[ \frac{a^2}{x^2} + \frac{a^2}{y^2} = 4 \]

### Final Answer
The locus of the midpoints of the tangents to the circle is:
\[ \frac{a^2}{x^2} + \frac{a^2}{y^2} = 4 \]