If the circles \( x^{2}+y^{2}+2 x+2 k y+6=0 \) and \( x^{2}+y^{2}+2 k y+k=0 \) intersect orthogonally, then k is
\[ \text{Since, the given circles intersect orthogonally.} \] \[ \therefore 2(1)(0) + 2(k)(k) = 6 + k \] \[ (\because 2g_1 g_2 + 2f_1 f_2 = c_1 + c_2) \] \[ \Rightarrow 2k^2 - k - 6 = 0 \] \[ \Rightarrow k = -\frac{3}{2}, 2 \]
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