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

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$x^{2} + y^{2} - 6 x + 4 = 0$

$x^{2} + y^{2} - 4 y + 2 = 0$

$x^{2} + y^{2} - 4 x + 5 = 0$

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

answer is B.

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Detailed Solution

Equation of circle passing through circles $x^{2} + y^{2} = 6 and x^{2} + y^{2} - 6 x + 8 = 0$

is $(x^{2} + y^{2} - 6) + λ (x^{2} + y^{2} - 6 x + 8) = 0,$

where $λ$ is a para meter and $λ$ ; $λ \in R$

Above circle passes through $(1, 1)$

$\begin{array}{l} ∴ (1 + 1 - 6) + λ (1 + 1 - 6 + 8) = 0 \\ - 4 + λ (4) = 04 λ = 4 \\ λ = 1 \end{array}$

$∴$ Required circle equation $(x^{2} + y^{2} - 6) + 1 (x^{2} + y^{2} - 6 x + 8) = 0$

$\begin{array}{l} 2 x^{2} + 2 y^{2} - 6 x + 2 = 0 \\ ∴ x^{2} + y^{2} - 3 x + 1 = 0 \end{array}$

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