Question

The locus of the mid-point of a chord of the circle $x^2 + y^2 = 4$ which subtends a right angle at the origin is:

• A. $x + y = 2$
• B. $x^2 + y^2 = 1$
• C. $x^2 + y^2 = 2$
• D. $x + y = 1$

Hint:

Understanding the perpendicularity condition simplifies locus problems.

Answer 1

The Correct Option is C

Let the mid-point of the chord be $(h,k)$. The perpendicular from the origin to the chord satisfies: \[ OC = \sqrt{h^2 + k^2}, \] Using trigonometry and given conditions, we derive: \[ h^2 + k^2 = 2 \]

Answer 2

Step 1: Given Information
The equation of the circle is \( x^2 + y^2 = 4 \).
Let the endpoints of the chord be \( A \) and \( B \), and the chord subtends a right angle at the origin \( O(0, 0) \).
Let \( M(h, k) \) be the midpoint of chord \( AB \). We are to find the locus of \( M \).

Step 2: Using Geometry and Vectors
Since the chord subtends a right angle at the origin, angle \( AOB = 90^\circ \).
This implies that triangle \( AOB \) is a right-angled triangle with the right angle at the origin.
According to the property of circles, if a chord subtends a right angle at the center, then the product of the vectors \( \vec{OA} \cdot \vec{OB} = 0 \) (as they are perpendicular).

Step 3: Coordinate Geometry Approach
Let the endpoints of the chord be \( A(x_1, y_1) \) and \( B(x_2, y_2) \), such that \( OA \perp OB \).
So \( \vec{OA} \cdot \vec{OB} = 0 \Rightarrow x_1x_2 + y_1y_2 = 0 \).
Now, the midpoint \( M(h, k) \) of \( AB \) is given by:
\[ h = \frac{x_1 + x_2}{2}, \quad k = \frac{y_1 + y_2}{2} \]
So, \( x_1 = 2h - x_2 \) and \( y_1 = 2k - y_2 \).
Substitute into the dot product condition:
\[ x_1x_2 + y_1y_2 = (2h - x_2)x_2 + (2k - y_2)y_2 = 0 \]
Expanding:
\[ 2hx_2 - x_2^2 + 2ky_2 - y_2^2 = 0 \Rightarrow 2hx_2 + 2ky_2 = x_2^2 + y_2^2 \]
Now, since both points lie on the circle \( x^2 + y^2 = 4 \), the coordinates of both points lie at a distance 2 from the origin. The chord lies within the circle, so the midpoint \( M(h, k) \) must satisfy a certain condition.
By a well-known result: the locus of the midpoint of a chord of a circle that subtends a right angle at the center is another circle with half the radius.
Given that the original circle has radius 2, the locus of midpoints will lie on a circle with radius \( \sqrt{2} \).

Step 4: Final Equation of the Locus
Thus, the required locus is:
\[ x^2 + y^2 = 2 \]
Conclusion:
The locus of the midpoint of the chord is the circle \( x^2 + y^2 = 2 \).