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 orthogonality, then $k$ is:

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Solution

We have,
$\begin{matrix} x^{2} + y^{2} + 2 x + 2 k y + 6 = 0 x^{2} + y^{2} + 2 k y + k = 0 2 \times 1 \times 0 + 2 \times k \times k = 6 + k 2 k^{2} = k + 6 2 k^{2} = k + 6 2 k^{2} - k - 6 = 0 2 k^{2} - 4 k + 3 k - 6 = 0 (2 k + 3) (k - 2) = 0 k = \frac{- 3}{2} k = 2 \end{matrix}$

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