The locus of mid point of a chord of the circle $x^{2} + y^{2} = 2$ which substends a right angle at the origin is

x^{2} + y^{2} = 3

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x + y = 2

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x^{2} + y^{2} = 1

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None of these

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Solution

The correct option is C $x^{2} + y^{2} = 1$
Let mid point be $P (h, k)$
and origin be $O (0, 0)$ which is also centre of the given circle.
Since chord making right angle at origin and $P$ is mid point, $O P$ will bisect the right angle.
$\Rightarrow O P = r cos 45^{\circ}$ where $r$ is radius of the given circle.
$\Rightarrow O P = \sqrt{(h - 0)^{2} + (k - 0)^{2}} = \sqrt{2} \times \frac{1}{\sqrt{2}} = 1$
Squaring we get, $h^{2} + k^{2} = 1$
Hence locus of $P (h, k)$ is, $x^{2} + y^{2} = 1$

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List- I	List- II
A. The locus of the point of intersection of the perpendicular tangents of $x^{2} + y^{2} = a^{2}$	1) $x^{2} + y^{2} = 5$ of
B. The locus of the mid-point of the chord of the circle $x^{2} + y^{2} = 8$ which is $\sqrt{2}$ units from origin is	2) $x^{2} + y^{2} = a^{2} {sec}^{2} α$
C. The locus of the point of intersection of the lines $x = \frac{1 - t^{2}}{1 + t^{2}}$ and $y = \frac{2 t}{1 + t^{2}}$	3) $x^{2} + y^{2} = 2 a^{2}$
D. The locus of the point whose chord of contact w.r.t $x^{2} + y^{2} = a^{2}$ making an angle '2 $α$ ' at the center is	4) $x^{2} + y^{2} = 2$ 5) $x^{2} + y^{2} = 1$

x^{2} + y^{2} + 2 x - 2 y - 2 = 0

90 °

The locus of the middle points of those chords of the circle $x^{2} + y^{2} = 4$ which subtend a right angle at the origin is

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