A variable chord of circle $x^{2} + y^{2} = 4$ is drawn from the point $P (3, 5)$ meeting the circle at the points $A$ and
$B$ . A point $Q$ is taken on this chord such that $2 P Q = P A + P B$ . Locus of $^{''} Q^{''}$ is

x^{2} + y^{2} + 3 x + 4 y = 0

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x^{2} + y^{2} = 36

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x^{2} + y^{2} = 16

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x^{2} + y^{2} - 3 x - 5 y = 0

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Solution

The correct option is D $x^{2} + y^{2} - 3 x - 5 y = 0$

Given, $2 P Q = P A + P B$
$\Rightarrow P Q - P A = P B - P Q$
$\Rightarrow A Q = B Q$

So, $Q$ is midpoint of $A B$
Perpendicular from centre bisects the chord, so
$O Q ⊥ A B$
$\Rightarrow$ Slope of $O Q \times$ Slope of $A B = - 1$

$\Rightarrow \frac{k - 0}{h - 0} \times \frac{k - 5}{h - 3} = - 1 \Rightarrow k^{2} - 5 k = - h^{2} + 3 h \Rightarrow k^{2} + h^{2} - 5 k - 3 h = 0$

Hence, the locus of $Q$ is
$x^{2} + y^{2} - 3 x - 5 y = 0$

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Similar questions

x^{2} + y^{2} = 4

P (3, 5)

A

B

Q

2 P Q = P A + P B

^{''} Q^{''}

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

From the origin chords are drawn to the circle $(x - 1)^{2} + y^{2} = 1$ . The equation of the locus of the middle points of these chords is

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

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

x^{2} + y^{2} = a^{2} .

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