JEE Main · 2006 · 30 Apr 2006 · Offline · Circle

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

\frac{2 π}{3}

at its centre is

Answer: (D)

Solution

Let

(h, k)

be the midpoint of a chord of the circle

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

that subtends an angle

\frac{2 π}{3}

at the centre. The radius drawn to the midpoint bisects both the chord and the central angle, so it makes

\frac{π}{3}

with the radius to an endpoint. In the right triangle formed by the centre, the midpoint and an endpoint,

cos \frac{π}{3} = \frac{distance from centre to midpoint}{radius}

, that is

\frac{1}{2} = \frac{h ^{2} + k ^{2}}{3}

. Hence

h^{2} + k^{2} = \frac{3}{2}

and

h^{2} + k^{2} = \frac{9}{4}

, so the midpoints lie on

x^{2} + y^{2} = \frac{9}{4}

JEE Main · 2006 · 30 Apr 2006 · Offline · Single correct option