The locus of the midpoints of chords of the circle ${ x }^{ 2 }+{ y }^{ 2 }-2x-2y-2=0$ which makes an angle $120^{0}$ at the center is
- ${ x }^{ 2 }+{ y }^{ 2 }-2x-2y+1=0$
- ${ x }^{ 2 }+{ y }^{ 2 }-x-y+1=0$
- ${ x }^{ 2 }+{ y }^{ 2 }-2x-2y-1=0$
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None of these
The circle is (x-1)^2 + (y-1)^2 = 4, center (1,1), radius r = 2. A chord subtending 120 degrees at the center has a distance d from the center given by d = r * cos(60) = 2 * 0.5 = 1. The locus of the midpoint (h,k) of a chord at distance d=1 from (1,1) is (h-1)^2 + (k-1)^2 = 1^2. Expanding: h^2 - 2h + 1 + k^2 - 2k + 1 = 1, so x^2 + y^2 - 2x - 2y + 1 = 0.
The centre of the given circle is at (1, 1) and the radius squared is 1 squared plus 1 squared minus negative 2, which is 4, giving a radius of 2. For a chord subtending 120 degrees at the centre, the perpendicular distance from the centre to the midpoint is found using the cosine of the half angle of 60 degrees, so d equals 2 times the cosine of 60 degrees, which is 1. The locus of the midpoints forms a circle with radius 1 around (1, 1), written as (x minus 1) squared plus (y minus 1) squared equals 1 squared. Expanding this equation gives x squared plus y squared minus 2x minus 2y plus 1 equals 0.