Q60 / 155 · WBJEE 2017 · Mathematics
The locus of the mid-points of the chords of the circle $x^{2}+y^{2}+2 x-2 y-2=0,$ which make an angle of $90^{\circ}$ at the centre is
- A$x^{2}+y^{2}-2 x-2 y=0$
- B$x^{2}+y^{2}-2 x+2 y=0$
- Correct answerC$x^{2}+y^{2}+2 x-2 y=0$
- D$x^{2}+y^{2}+2 x-2 y-1=0$
Solution
Given, equation of circle is $x^{2}+y^{2}+2 x-2 y-2=0$
$\Rightarrow \quad(x+1)^{2}+(y-1)^{2}=4$
$\therefore$ Centre (-1,1) and radius $=2$
Let $(h, k)$ be the mid-point of chord.
From figure,
$$
O P=\sqrt{(h+1)^{2}+(k-1)^{2}}
$$
$\ln \Delta O A P$
$$
\begin{aligned}
\sin 45^{\circ}=& \frac{O P}{O A} \\
\Rightarrow \quad \frac{1}{\sqrt{2}}=\frac{\sqrt{(h+1)^{2}+(k-1)^{2}}}{2}
\end{aligned}
$$
On squaring both sides, we get
$$
(h+1)^{2}+(k-1)^{2}=2
$$
$\Rightarrow \quad h^{2}+k^{2}+2 h-2 k=0$
$\therefore$ Locus of $P$ will be
$$
x^{2}+y^{2}+2 x-2 y=0
$$
From figure,
$$
O P=\sqrt{(h+1)^{2}+(k-1)^{2}}
$$
$\ln \Delta O A P$
$$
\begin{aligned}
\sin 45^{\circ}=& \frac{O P}{O A} \\
\Rightarrow \quad \frac{1}{\sqrt{2}}=\frac{\sqrt{(h+1)^{2}+(k-1)^{2}}}{2}
\end{aligned}
$$
On squaring both sides, we get
$$
(h+1)^{2}+(k-1)^{2}=2
$$
$\Rightarrow \quad h^{2}+k^{2}+2 h-2 k=0$
$\therefore$ Locus of $P$ will be
$$
x^{2}+y^{2}+2 x-2 y=0
$$Hence, the correct answer is option: 'C'.