The locus of the mid-points of the chords of the circle $x^{2}+y^{2}-4x-2y-11=0$which subtend $60^{0}$ at the centre is
- $x^{2}+y^{2}-4x-2y-7=0$
- $x^{2}+y^{2}+4x+2y-7=0$
- $x^{2}+y^{2}-2x-4y-7=0$
- $x^{2}+y^{2}+2x+4y+7=0$
The chord subtends 60 degrees at the center, so the triangle formed by the center and the chord endpoints is equilateral. The distance from the center (2,1) to the midpoint of the chord is r * cos(30) = sqrt(16) * sqrt(3)/2 = 2 * sqrt(3). The locus of midpoints is a circle concentric with the original circle, with radius equal to this distance. Radius squared = 12. Equation: (x-2)^2 + (y-1)^2 = 12. x^2 - 4x + 4 + y^2 - 2y + 1 = 12. x^2 + y^2 - 4x - 2y - 7 = 0.
The given circle equation is x squared plus y squared minus 4x minus 2y minus 11 equals 0, which has its centre at (2, 1) and a radius of 4. The midpoint of a chord lies on a smaller concentric circle, and its distance from the centre is calculated using the formula r cosine of half the angle. For an angle of 60 degrees, the distance from the centre to the midpoint is 4 times cosine of 30 degrees, which simplifies to 2 times the square root of 3. The locus of these midpoints is a circle with radius 2 root 3 centred at (2, 1), given by (x - 2) squared plus (y - 1) squared equals 12; expanding this yields the equation x squared + y squared - 4x - 2y - 7 = 0.