Question

The locus of midpoint of chord of the circle $x^2+y^2-2x-2y-2=0$, which makes an angle of ${120}^{\circ }$ at the centre, is

• A. $x^2+y^2-x-y+3=0$
• B. $x^2+y^2-2x-2y+4=0$
• C. $x^2+y^2-4x-4y+8=0$
• D. $x^2+y^2-2x-2y+1=0$

Answer 1

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

For the given circle $x^2+y^2-2x-2y-2=0$
Centre, $C\equiv (1,1)$
and radius, $r=\sqrt{1^2+1^2-(-2)}=2$

From the above figure in $△AOM$
$\cos 60°=\frac{OM}{r}\Rightarrow OM=1$
But $OM=\sqrt{(h-1{)}^2+(k-1{)}^2}$
$\therefore (h-1{)}^2+(k-1{)}^2=1$
Hence, the locus of $(h,k)$ is $(x-1{)}^2+(y-1{)}^2=1$
or, $x^2+y^2-2x-2y+1=0.$