Let C 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 C that subtend an a - Mathematics MCQ AIEEE 2006 2026 | ExamTest.live

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Let C 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 C that subtend an angle of 2 π /3 at its centre is:

Solution & Step-by-step Explanation

Let the midpoint of the chord be M (h, k) .The distance of the midpoint from the center (0, 0) is OM = h^{2} + k^{2} .In the triangle formed by the center O, midpoint M, and one end of the chord A : O A = radius = 3 . ∠ A OM = \frac{1}{2} (angle subtended) = \frac{1}{2} (2 π /3) = π /3 .In △ O A M : cos (π /3) = \frac{OM}{O A} = \frac{h ^{2} + k ^{2}}{3} 1/2 = \frac{h ^{2} + k ^{2}}{3} ⟹ h^{2} + k^{2} = 3/2 Squaring both sides: h^{2} + k^{2} = 9/4 .The locus is x^{2} + y^{2} = 9/4 .

Try it yourself before checking the explanation above.

Let C 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 C that subtend an angle of 2 π /3 at its centre is:

x^{2} + y^{2} = 3/2

x^{2} + y^{2} = 1

x^{2} + y^{2} = 27/4

x^{2} + y^{2} = 9/4

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