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 $90º$ 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

Answer: (C)

Given, equation of circle is
$x^2+y^2+2x-2y-2=0$
$\Rightarrow (x+1{)}^2+(y-1{)}^2=4$
$\therefore$ Centre $(-1,1)$ and radius $=2$
Let $(h,k)$ be the mid-point of chord.

From figure,
$OP=\sqrt{(h+1{)}^2+(k-1{)}^2}$
In $△OAP$,
$\sin 45^{\circ }=\frac{OP}{OA}$
$\Rightarrow \frac{1}{\sqrt{2}}=\frac{\sqrt{(h+1{)}^2+(k-1{)}^2}}{2}$
On squaring both sides, we get
$(h+1{)}^2+(k-1{)}^2=2$
$\Rightarrow h^2+k^2+2h-2k=0$
$\therefore$ Locus of $P$ will be
$x^2+y^2+2x-2y=0$