The locus of the mid point of the chord of the circle $x^{2} + y^{2} - 2 x - 2 y - 2 = 0$ which makes an angle of $120^{0}$ at the centre is.

x^{2} + y^{2} - 2 x - 2 y + 1 = 0

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x^{2} + y^{2} + x + y + 1 = 0

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x^{2} + y^{2} - 2 x - 2 y + 1 = 0

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x^{2} + y^{2} + x - y - 1 = 0

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Solution

The correct option is A $x^{2} + y^{2} - 2 x - 2 y + 1 = 0$
Given circle is $x^{2} + y^{2} - φ - φ - 2 = 0 C e n t r e (1, 1) R a d i u s = 2 I n Δ O B P C o s 60^{\circ} = \frac{O P}{O B} O P = 1 (x - 1)^{2} + (y - 1)^{2} = 1 x^{2} + 1 + y^{2} - φ + 1 = 1 x^{2} + y^{2} - 2 x - φ + 1 = 0$

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Similar questions

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

90 °

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

120^{\circ}

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

120^{\circ}

x^{2} + y^{2} - 2 x - 2 y + 1 = 0

From origin, chords are drawn to the circle $x^{2} + y^{2} - 2 y = 0$ .The locus of the middle points of these chords is

P (2, 2)

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

A

B .

△ P A B

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