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 $90�$ at the centre is
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 $90�$ at the centre is
Updated On: Feb 2, 2024
- $x^2 + y^2 - 2x - 2y = 0$
- $x^2 + y^2 - 2x + 2y = 0$
- $x^2 + y^2 + 2x - 2y = 0$
- $x^2 + y^2 + 2x - 2y - 1=0$
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The Correct Option is C
Solution and Explanation
Given, equation of circle is
$x^{2}+y^{2}+2 x-2 y-2=0$
$\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,
$O P=\sqrt{(h+1)^{2}+(k-1)^{2}}$
In $\triangle O A P$,
$\sin 45^{\circ}=\frac{O P}{O A}$
$\Rightarrow \frac{1}{\sqrt{2}}=\frac{\sqrt{(h+1)^{2}+(k-1)^{2}}}{2}$
On squaring both sides, we get
$ (h+1)^{2}+(k-1)^{2}=2 $
$\Rightarrow h^{2}+k^{2}+2 h-2 k=0$
$\therefore$ Locus of $P$ will be
$x^{2}+y^{2}+2 x-2 y=0$
$x^{2}+y^{2}+2 x-2 y-2=0$
$\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,
$O P=\sqrt{(h+1)^{2}+(k-1)^{2}}$
In $\triangle O A P$,
$\sin 45^{\circ}=\frac{O P}{O A}$
$\Rightarrow \frac{1}{\sqrt{2}}=\frac{\sqrt{(h+1)^{2}+(k-1)^{2}}}{2}$
On squaring both sides, we get
$ (h+1)^{2}+(k-1)^{2}=2 $
$\Rightarrow h^{2}+k^{2}+2 h-2 k=0$
$\therefore$ Locus of $P$ will be
$x^{2}+y^{2}+2 x-2 y=0$
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