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Find the locus of the mid point of the chord of the circle $x^2+y^2-2x-2y-2=0$ which makes an angle of $120^{\circ} $ at the centre.

My attempt:

Given equation of circle is $$x^2+y^2-2x-2y-2=0$$ Center$=(-g,-f)=(1,1)$ Radius $r=\sqrt {g^2+f^2-c}$ Let $AB$ bw a chord and $P$ be it's mid point. If $C$ is the centre of the circle then $\angle ACB=120^{\circ}$ So, $\angle ACP=60^{\circ}$

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    $\begingroup$ Since the $CA$ (the radius), $\angle ACP$, and $\angle APC=90$ are constant, then $CP$ is constant. Therefore the locus is a circle with center $C$. The radius $CP$ you can find using Pythagoras or trigonometry to be $1$. This is enough to write the equation: $(x-1)^2+(y-1)^2=1$. $\endgroup$ – logarithm May 29 at 14:09
  • $\begingroup$ hint: $\triangle{ACP}$ is right 30-60-90 triangle so it's going to be a circle of radius $1$ centered at $C$ $\endgroup$ – Vasya May 29 at 14:24
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The given circle has a radius of $2$ and the center of $(1,1)$

The distance from the center of the given circle to the midpoint of cords is a constant of $1$ due to the fact that in the right triangle formed by the center, the midpoint and one end of the cord we have a $30$ degree angle opposite to the segment connecting the center to the midpoint.

Therefore the locus is a circle with the same center and the radius $1$ that is $$(x-1)^2 + (y-1)^2 =1$$

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Clearly all such chords are the same length.

Any rotation around the origin that sends one chord to another must send the midpoint of the first to the second, so we can obtain all midpoints by rotating one around the centre, implying the locus is a circle with the same centre as the original circle.

We now need to find one midpoint to identify the radius. Choosing a vertical chord here makes calculation easier.

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