Question:medium

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

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For any circle \(x^2 + y^2 = r^2\), the locus of the midpoint of a chord subtending an angle \(\theta\) at the center is always a circle \(x^2 + y^2 = (r \cos \frac{\theta}{2})^2\).
Updated On: Apr 19, 2026
  • \( x + y = 2 \)
  • \( x^2 + y^2 = 1 \)
  • \( x^2 + y^2 = 2 \)
  • \( x + y = 1 \)
Show Solution

The Correct Option is C

Solution and Explanation

Step 1: Understanding the Concept: 
We can solve this using either simple coordinate geometry properties (distance from center) or the homogenization method. Using the geometric properties of the circle is the most straightforward and fastest method. 
Step 2: Key Formula or Approach: 
For a circle centered at the origin, a chord that subtends a specific angle $2\theta$ at the center creates an isosceles triangle with the two radii. The perpendicular distance $d$ from the center to the chord (which is also the midpoint) is given by: \[ d = r \cos \theta \] Step 3: Detailed Explanation: 
The circle is $x^2 + y^2 = 4$. Its center is $(0, 0)$ and its radius is $r = 2$. 
Let the midpoint of the chord be $(h, k)$. 
The distance from the center $(0,0)$ to the midpoint $(h, k)$ is $d = \sqrt{h^2 + k^2}$. 
The chord subtends an angle of $90^\circ$ at the origin (center). The line joining the center to the midpoint bisects this angle. 
Therefore, the angle $\theta$ between the radius and the perpendicular bisector is $90^\circ / 2 = 45^\circ$. 
Using the relation from the right-angled triangle formed: \[ d = r \cos(45^\circ) \] Substitute $d$ and $r$: \[ \sqrt{h^2 + k^2} = 2 \left( \frac{1}{\sqrt{2}} \right) = \sqrt{2} \] Squaring both sides to find the locus: \[ h^2 + k^2 = 2 \] Replacing $(h, k)$ with general coordinates $(x, y)$, the equation of the locus is: \[ x^2 + y^2 = 2 \] Step 4: Final Answer: 
The locus of the midpoint is $x^2 + y^2 = 2$.

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