The circle which passes through the points of intersection of the circles $x^{2} + y^{2} - 6 x + 8 = 0, x^{2} + y^{2} = 6$ and the point $(1, 1)$ is

x^{2} + y^{2} + 3 x + 1 = 0

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x^{2} - y^{2} - 3 x + 1 = 0

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x^{2} + y^{2} - 3 x + 1 = 0

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x^{2} - y^{2} + 3 x + 1 = 0

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Solution

The correct option is C $x^{2} + y^{2} - 3 x + 1 = 0$
The circle is $S + λ S^{'} = 0$
$\Rightarrow x^{2} + y^{2} - 6 x + 8 + λ (x^{2} + y^{2} - 6) = 0$
It passes through $(1, 1)$
$∴ 1 + 1 - 6 + 8 + λ (1 + 1 - 6) = 0$
$\Rightarrow λ = 1$
The circle is $x^{2} + y^{2} - 6 x + 8 + x^{2} + y^{2} - 6 = 0$
or $2 x^{2} + 2 y^{2} - 6 x + 2 = 0$
or $x^{2} + y^{2} - 3 x + 1 = 0$

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