Circles
Locus of midpoint of chord
Grade None
Question:
<p>Let <em>C</em> be the circle with centre (0, 0) and radius 3 units. The equation of the locus of the mid points of the chords of the circle <em>C</em> that subtend an angle of \(2\pi/3\) at its centre is</p>
<p>\(x^2 + y^2 = \dfrac{3}{2}\)</p>
<p>\(x^2 + y^2 = 1\)</p>
<p>\(x^2 + y^2 = \dfrac{27}{4}\)</p>
<p>\(x^2 + y^2 = \dfrac{9}{4}\)</p>
Step-by-Step Solution
Key Concept: For a chord subtending angle 2θ at center, if M is the midpoint, then OM ⊥ chord and OM = r·cos(θ). Here θ = π/3, so the locus is a circle with radius 3·cos(π/3) = 3/2.
<p><strong>Step 1:</strong> For a chord of circle C (center O, radius r=3) subtending angle 2π/3 at center, let M be the midpoint of the chord.</p><p><strong>Step 2:</strong> Since M is the midpoint of chord AB, OM ⊥ AB. In triangle OAM (where A is an endpoint), ∠AOM = (2π/3)/2 = π/3.</p><p><strong>Step 3:</strong> Using right triangle OAM: OM = OA·cos(π/3) = 3·cos(π/3) = 3·(1/2) = 3/2.</p><p><strong>Step 4:</strong> Since OM is constant for all such chords, M lies on a circle with center O(0,0) and radius 3/2.</p><p><strong>Step 5:</strong> The locus equation is x² + y² = (3/2)² = 9/4.</p><p>∴ Answer: <strong>x² + y² = 9/4</strong> (or equivalent form)</p>
Correct Answer: D