Show that the locus of the mid-points of the chords of the circle $x^{2} + y^{2} - 2 x - 2 y - 2 = 0$ which make an angle of $120^{\circ}$ at the centre is $x^{2} + y^{2} - 2 x - 2 y + 1 = 0$ .

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Solution

The centre of the given circle is $(3 / 2, - 1 / 2)$ and its radius is $3 / 2$ . From the figure if $M (h, k)$ be the middle point of chord $A B$ subtending an angle $2 π / 3$ at $C$ , then
$\frac{C M}{A C} = cos \frac{π}{3} = \frac{1}{2}$ or $4 C M^{2} = A C^{2}$
or $4 [(h - 3 / 2)^{2} + (k + 1 / 2)^{2}] = 9 / 4$
$∴$ Locus is $4 x^{2} + 4 y^{2} - 12 x + 4 y + (31 / 4) = 0$ .

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