The locus of the midpoint of chord of the circle `x^(2)+y^(2)=1` which subtends a right angle at the origin is

Options

A. `x^(2)+y^(2)=1/4`
B. `x^(2)+y^(2)=1/2`
C. `xy=0`
D. `x^(2)-y^(2)=0`

Answer

To find the locus of the midpoint of a chord of the circle \( x^2 + y^2 = 1 \) that subtends a right angle at the origin, we can follow these steps:

### Step 1: Understand the Circle and Chord
The equation \( x^2 + y^2 = 1 \) represents a circle centered at the origin (0, 0) with a radius of 1. A chord of this circle subtending a right angle at the origin means that the angle formed by the lines connecting the origin to the endpoints of the chord is \( 90^\circ \).

### Step 2: Define the Midpoint
Let the endpoints of the chord be \( A(a_1, b_1) \) and \( B(a_2, b_2) \). The midpoint \( C(h, k) \) of the chord \( AB \) can be expressed as:
\[ C(h, k) = \left( \frac{a_1 + a_2}{2}, \frac{b_1 + b_2}{2} \right) \]

### Step 3: Use the Right Angle Condition
Since the chord subtends a right angle at the origin, we can use the property of the circle that states that if a chord subtends a right angle at the center, then the product of the distances from the origin to the endpoints of the chord is equal to the square of the radius of the circle. Thus, we have:
\[ OA \cdot OB = r^2 \]
Where \( OA \) and \( OB \) are the distances from the origin to points \( A \) and \( B \), respectively. Since the radius \( r = 1 \):
\[ OA^2 + OB^2 = 1^2 \]

### Step 4: Express Distances
The distances \( OA \) and \( OB \) can be expressed as:
\[ OA^2 = a_1^2 + b_1^2 \quad \text{and} \quad OB^2 = a_2^2 + b_2^2 \]
Since both points lie on the circle, we have:
\[ a_1^2 + b_1^2 = 1 \quad \text{and} \quad a_2^2 + b_2^2 = 1 \]

### Step 5: Use the Midpoint Formula
Using the midpoint \( C(h, k) \):
\[ h = \frac{a_1 + a_2}{2}, \quad k = \frac{b_1 + b_2}{2} \]

### Step 6: Relate Midpoint to Circle Equation
To find the locus of the midpoint, we can express \( a_1 \) and \( a_2 \) in terms of \( h \) and \( k \):
\[ a_1 + a_2 = 2h \quad \text{and} \quad b_1 + b_2 = 2k \]

### Step 7: Use the Right Angle Condition
From the right angle condition, we have:
\[ a_1 a_2 + b_1 b_2 = 0 \]
Using the identities:
\[ (a_1 + a_2)^2 = a_1^2 + a_2^2 + 2a_1a_2 \quad \text{and} \quad (b_1 + b_2)^2 = b_1^2 + b_2^2 + 2b_1b_2 \]
We can substitute \( a_1^2 + b_1^2 = 1 \) and \( a_2^2 + b_2^2 = 1 \) into the equations.

### Step 8: Final Equation
After substituting and simplifying, we find:
\[ h^2 + k^2 = \frac{1}{2} \]
This represents a circle with center at the origin and radius \( \frac{1}{\sqrt{2}} \).

### Conclusion
Thus, the locus of the midpoint of the chord of the circle \( x^2 + y^2 = 1 \) which subtends a right angle at the origin is given by:
\[ x^2 + y^2 = \frac{1}{2} \]