If the circle $x^{2} + y^{2} + 2 x + 2 k y + 6 = 0$ and $x^{2} + y^{2} + 2 k y + k = 0$ intersect orthogonally then $k$ is

2, - \frac{3}{2}

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- 2

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- \frac{3}{2}

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\frac{3}{2}

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Solution

The correct option is B $2, - \frac{3}{2}$
we know that the condition of orthogonally of two circles is-

$2 g_{1} g_{2} + 2 f_{1} f_{2} = c_{1} + c_{2}$

$∴ 2 (1) (0) + 2 (k) (k) = 6 + k$

$2 k^{2} - k - 6 = 0$

$2 k^{2} - 4 k + 3 k - 6$ = $0$

$2 k (k - 2) + 3 (k - 2) = 0$

$k = 2$ or $k = \frac{- 3}{2}$

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