JEE Advance - Mathematics (1993 - No. 16)
The equation of the locus of the mid-points of the circle $$4{x^2} + 4{y^2} - 12x + 4y + 1 = 0$$ that subtend an angle of $$2\pi /3$$ at its centre is.................................
$$16{x^2} + 16{y^2} - 48x + 16y + 31 = 0$$
$$16{x^2} + 16{y^2} + 48x - 16y + 31 = 0$$
$$16{x^2} - 16{y^2} - 48x + 16y + 31 = 0$$
$$16{x^2} + 16{y^2} - 48x - 16y + 31 = 0$$
$$16{x^2} + 16{y^2} - 48x + 16y - 31 = 0$$