The locus of the centre of the circle which bisects the circumferences of the circles $x^{2} + y^{2} = 4$ & $x^{2} + y^{2} - 2 x + 6 y + 1 = 0$ is

a straight line

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a circle

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a parabola

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pair of straight line

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Solution

The correct option is A a straight line
Given circles are, $x^{2} + y^{2} = 4$ and $x^{2} + y^{2} - 2 x + 6 y + 1 = 0$

The locus of the center of the circle bisects the circumferences of these two circles.

$∴$ Radical axis, $S_{1} - S_{2} = 0$

$⟹ (x^{2} + y^{2} - 4) - (x^{2} + y^{2} - 2 x + 6 y + 1) = 0$

$2 x - 6 y - 5 = 0$ . This is a straight line.

The center lies on the line parallel to the above line.

$⟹ 2 x - 6 y + λ = 0$

This is a straight line.

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