Question

The locus of the mid-points of the chords of the circle $x^2+y^2+2x-2y-2=0$ which make an angle of ${90}^{\circ }$ at the centre is

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

Answer 1

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

Given circle is $x^2+y^2+2x-2y-2=0$ which can be written as $(x+1{)}^2+(y-1{)}^2=4$

So, centre of the circle is $(-1,1)$ and radius is $2$

Let $O$ be the centre and $AB$ be the chord of the circle such that it makes an angle of ${90}^o$ at the centre

Draw $OA$ and $OB$

$\therefore \angle AOB={90}^o$

Now, $OA$ and $OB$ are radius of the circle

$\therefore OA=OB$

$⟹\angle OAB=\angle OBA$ ....... [Angle opposite to equal sides are equal]

But, $\angle OAB+\angle OBA+\angle AOB={180}^o$

$⟹2\angle OAB={180}^o-{90}^o$

$\angle OAB={45}^o=\angle OBA$

$OP$ is the perpendicular drawn from the centre of the circle

$\therefore p(h,k)$ is the midpoint of the chord $AB$

Join $OP$ which makes an angle of ${90}^o$ with the chord $AB$.

Now, $△APO$ is a right angled triangle such that $\angle APO={90}^o$

$\therefore \sin {45}^o=\frac{OP}{OA}=\frac{OP}{2}\Rightarrow OP=\sqrt{2}$.

But, $OP=\sqrt{(h+1{)}^2+(k-1{)}^2}$

$\therefore \sqrt{(h+1{)}^2+(k-1{)}^2}=\sqrt{2}$

$⟹h^2+2h+1+k^2-2k+1=2$

$⟹h^2+k^2+2h-2k=0$

$\therefore$ Locus of $P$ is $x^2+y^2+2x-2y=0$