# Find the locus of the mid point of the chord

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, $$\angle ACP=60^{\circ}$$

• 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: $(x-1)^2+(y-1)^2=1$. – logarithm May 29 at 14:09
• hint: $\triangle{ACP}$ is right 30-60-90 triangle so it's going to be a circle of radius $1$ centered at $C$ – Vasya May 29 at 14:24

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 $$(x-1)^2 + (y-1)^2 =1$$