Question

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

• A. $x^2+y^2-2x-2y+1=0$
• B. $x^2+y^2-x-y+1=0$
• C. $x^2+y^2-2x-2y-1=0$
• D. None of these

Answer 1

The correct option is A $x^2+y^2-2x-2y+1=0$

Given equation of circle is $x^2+y^2-2x-2y-2=0$

Let mid point of the chord $AB$ is $(h,k)$

Its center is $(1,1)$ and radius $=\sqrt{1+1+2}=2=OB$.

In $△OPB,\angle OBP={30}^0$

$\therefore \sin {30}^0=\frac{OP}{2}\Rightarrow OP=1$

Since $OP=1\Rightarrow {(h-1)}^2+{(k-1)}^2=1$

$\Rightarrow h^2+k^2-2h-2k+1=0$

$\therefore$ Locus of the mid point of chord is

$x^2+y^2-2x-2y+1=0$