The locus of the mid -point of the chords of a circle `x^(2)+y^(2)=4` ,which subtends a right angle at the centre , is

Options

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

Answer

To find the locus of the mid-point of the chords of the circle \(x^2 + y^2 = 4\) that subtend a right angle at the center, we can follow these steps:

### Step 1: Understanding 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 \(r = 2\).

### Step 2: Midpoint of the Chord
Let the midpoint of the chord be \(P(h, k)\). Since the chord subtends a right angle at the center, we can use the property that the distance from the center of the circle to the midpoint of the chord is given by:
\[ CP = r \cdot \sin(\theta) \]
where \(\theta\) is half the angle subtended by the chord at the center. For a right angle, \(\theta = 45^\circ\).

### Step 3: Calculate the Distance
Since \(CP\) is the distance from the center to the midpoint \(P(h, k)\), we can express it as:
\[ CP = \sqrt{h^2 + k^2} \]
Given that \(r = 2\) and \(\sin(45^\circ) = \frac{1}{\sqrt{2}}\), we have:
\[ CP = 2 \cdot \sin(45^\circ) = 2 \cdot \frac{1}{\sqrt{2}} = \sqrt{2} \]

### Step 4: Set Up the Equation
Equating the two expressions for \(CP\):
\[ \sqrt{h^2 + k^2} = \sqrt{2} \]
Squaring both sides gives:
\[ h^2 + k^2 = 2 \]

### Step 5: Replace Variables
Now, we can replace \(h\) and \(k\) with \(x\) and \(y\) respectively:
\[ x^2 + y^2 = 2 \]

### Conclusion
Thus, the locus of the mid-point of the chords of the circle \(x^2 + y^2 = 4\) that subtend a right angle at the center is given by:
\[ \boxed{x^2 + y^2 = 2} \]